Work And Energy - Class 9 Physics - Extra Questions

a) How are work, force and displacement related.
b) Find the work done by a pulley when it lifts a block through 5 m off the ground with a 10 N force.

a) Work is directly related to both the force applied to the object and the displacement the object moves. Work can be calculated with this equation: $$Work = Force\times Displacement .$$

While force and displacement are inversely propotional. The equation showing the relationship is $$W= Fd$$.
b) $$W= Fd $$
$$F=10N$$
$$d=5m$$
$$  W = 10N \times 5m = 50J$$


You lift a bag of rice of  450N  weight  to height 1.2m and carry it a distance of 10m to the kitchen. How much work is done?

The work is done only in lifting the bag to the height 1.2m, 
Work is given by
$$W=F \times s$$     $$F$$ is force $$s$$ displacement in direction of force.
$$=450\times 1.2=540J$$

and $$no\ work\ is\ done\ in\ carrying\ it\ to\ the\ kitchen\ $$, since in that case displacement is perpendicular to the force of gravity. Because displacement is horizontal to ground but force of gravity is downwards so both perpendicular to each other.


Calculate the work done in pushing a cart through a distance of $$10m$$ when the force applied on it is equal to $$120N$$.

Work done =Force $$\times$$ Displacement
                   $$=120\times 10$$
                   $$=1200 J$$


A force of $$500\ dyne$$ acts on an object where the object moves through $$8\ m$$ in the direction of force. Calcualate the work done in this case

$$F=500\ dyne=500\times 10^{-5}=0.005N$$
$$s=8m$$
$$Workdone=Force\times Distance$$
$$W=0.005\times 8$$
$$W=0.04J$$


If a ski lift raises $$100$$ passengers averaging $$660 N$$ in weight to a height of $$150 m$$ in $$60.0 s$$, at constant speed, what average power is required of the force making the lift?

Given,
Number of passengers $$=100$$
Averaging weight $$=660\ N$$
Height to left $$h=150\ m$$ 
time required $$t=60\ s$$
The total weight is $$(100)(660 \ N) = 6.60 × 10^4 N$$, 

$$P = \dfrac{W}{t} = \dfrac{F \times s}{t} = \dfrac{6.60 × 10^4 \times 150 }{60} = 1.65 × 10^5 \ W$$.


A man rowing a boat upstream is at rest with respect to the shore. Is he doing work?

Yes, the boat is doing work against the river current but there is no work done by boat against shore because there is no displacement taking place with respect to shore.


The time it takes to move an object at a distance of $$5 \ m$$ is $$10 \ s$$. If the force applied is $$30\ N$$, what is the power?

Work done is defined as $$W=Fs$$, where $$F$$ is the force applied and $$s$$ is the displacement in the direction of force.

Given, $$F=30 \ N$$, and $$s = 5\ m$$
$$W = F\cdot s = 30\times 5=150 \ J$$

Power is defined as the rate of work done $$P = \dfrac{W}{t}$$
Given, $$t=10sec$$

hence, power $$  P = \dfrac{150}{10}=15\ W$$


The $$S.I. $$ unit of power is _____

$$S.I. $$ unit of power is watt, which is equal to joule per second ie $$j/s$$.


The potential energy of a body is due to its ______ and kinetic energy of body is due to its ______.

The potential energy of a body is due to its state of rest or position and kinetic energy of body is due to its state of motion.


Name the type of energy (kinetic or potential) possessed by the following:
A moving football

Energy posses by an object due to its motion is known as Kinetic energy. Whereas energy posses by an object due to its posion in the gravitational field is knon as potential energy. Here football is moving hence it will have Kinetic energy


What is potential energy? State its unit.

The energy possessed by a body due to its change in position or shape is called the potential energy. The SI unit of potential energy is $$Joule$$.


Give reason for the following-
No work is done if a man is pushing against a wall.

We know,
$$Work\ done=Force\times displacement$$
If a man is pushing against a wall, the wall does not move from its place. Therefore, displacement is zero.
Hence, the work done is zero.


Name the type of energy (kinetic or potential) possessed by the following:
A bullet fired from a gun.

A bullet fired from a gun possesses kinetic energy due to its motion.


State the $$S.I. $$ unit of work and define it.

$$ S.I. $$ unit of work is $$Joule$$.

$$1$$ Joule is the amount of work done when a force of $$1 \ N$$  displaces the body through a distance of $$1 \ m$$ in the direction of force.



Name the type of energy (kinetic or potential) possessed by the following:
A moving bus.

A moving bus possesses kinetic energy by virtue of its motion.


A pump raises water by spending $$  4 \times  10^5 \  J $$ of energy in $$ 10 \ s. $$ Find the power of pump.

Energy spent  $$   E  = 4 \times 10^5 \ J $$
time taken $$ = 10 \ s $$

$$Power  = \dfrac{Energy}{Time}$$
$$ P = \dfrac { 4 \times 10^5}{10} = 4 \times 10^4 \ W $$


Give reason for the following.
A teacher moving around in the class is doing work but a child standing and reading a book is not doing any work.

A child reading a book while standing is not moving from its place i.e. displacement is zero. 
$$ W  = F \times S $$
$$ W  = F \times O = O $$ 
Hence, child is not doing any work. 
But teacher is in motion and therefore doing work.


Define work. Is work a scalar or a vector?

Work is said to be done only when the force applied on a body makes the body move. It is a scalar quantity. 


Give reason for the following.
A horse and a dog are running with the same speed. Which one of them has more kinetic energy than the other.

K.E  = $$\dfrac{1}{2} mv ^{2}$$

A horse has more mass than dog. As both are running with the same speed. $$ M_1 $$ of horse is greater than $$ M_2 $$ of dog.
$$ \therefore K.E. $$ of horse is more than $$ K.E. $$ of dog.


The displacement is opposite to the frictional force, is the work done by the frictional force positive or negative? 

If the displacement is in the opposite direction work done by the frictional force is negative.
$$\text{Work = Force} \times \text{displacement in the force direction}$$


A ball of mass 0.4 kg is thrown vertically upward with a velocity 14 m /s . Calculate its kinetic energy and potential energy after 1 s


1879729_1849688_ans_9ad30e9918654a60bf80ebbde56054c4.jpg


What is work? Mention the factors on which the work done by a force depends?

Work is said to be done when the point of application of a force moves.
$$Work=Force\times distance$$
$$ W = F \times d $$ 
The work done by the force on a body depends on two factors.
  1. The magnitude of the force and 
  2. distance that the body moves


A man pushes a vehicle by sitting inside the vehicle. The vehicle does not move. While when he came outside the vehicle and pushed the vehicle moved. Why?

When the man is sitting inside the force acting is internal. To move the vehicle an external unbalanced force is required. When he comes outside an external unbalanced force is applied and the vehicle moves. 


What will be the result when a man tries to move a vehicle by punching it, standing inside the vehicle? The vehicle moves the vehicle doesn't move.

The vehicle doesn't move.
The punch will work as an internal force, which does not give acceleration i.e. cause movement. Only external force cause movement.


$$1$$ Horse power $$=$$ _____ watt.

$$1$$ Horse power (HP) = 746 watt.


When does a force do work?

Work is said to be done only when the force applied on a body makes the body moves (i.e., there is displacement of the body).


Can kinetic energy of a system be increased or decreased without applying any external force on the system? (Yes/No)

Yes, the kinetic energy of a body can be increased or decreased without the application of an external force. 

Example: rocket flying in space gains or loses speed by ejecting high speed gases.



Give two examples where a body possesses both kinetic energy as well as potential energy.

Following are the two examples where a body possess both kinetic energy as well as potential energy:
a) A flying airplane
b) A man climbing stairs


The blades of a windmill sweep out a circle of area A.
(a) If the wind flows at a velocity v perpendicular to the circle, what is the mass of the air passing through it in time t?
(b) What is the kinetic energy of the air?
(c) Assume that the windmill converts 25% of the winds energy into electrical energy, and that A = 30 $$m^2$$, v = 36 km/h and the density of air is 1.2 kg $$m^{3}$$. What is the electrical power produced?

(a) 
Density of air $$= \rho \ \ kg /m^3$$
Time taken = $$t$$ sec
Velocity of air = $$v$$ m/s
Mass of the air = density $$\times $$ volume $$= \rho A v t$$
(b) 
$$KE = \cfrac{1}{2}mv^2$$
    $$ = \cfrac{1}{2}(\rho A vt) v^2$$
    $$ = \cfrac{1}{2}(\rho A v^3t) $$
(c) 
Velocity, v= 36 km/hr = 10 m/s
$$\rho = 1.2 kg/m^3$$
$$A = 30 m^2$$
Work done = change in kinetic energy
$$ = \dfrac{1}{2} \times 1.2 \times 30 \times 10^3 t \,Nm$$
$$ = 18000t \,J$$
Power = $$work /t = 18000 W$$
Efficiency = 25%
Actual power$$ = 18000 \times 25/100 = 4500 W$$

379722_419857_ans_96778b2ca20148659c65149e90a0b0c8.PNG


The SI unit of work and energy is Joule.
True=1; False=0

The SI unit of work and energy is Joule(J) which is defined as the work experienced by a force of one Newton through a displacement of one metre.


A man pushing against a wall does not do any work, explain briefly.

Work done (W) = force $$\ast$$ displacement.
Hence, work is said to be done when there is displacement. In this case the wall does not displace from its position even though the force is applied and since displacement is zero work done is said to be zero.


A person holds a bundle of hay over his head for $$30$$ minutes and gets tired. Has he done some work or not? Justify your answer.


2170364_464564_ans_b421d7dc16294a5e9cef467659e0b9a6.jpeg


What is the work done by the force of gravity on a satellite moving round the earth? Justify your answer.

Hint: Use the concept of work done  $$W= \overrightarrow {F}. \overrightarrow {S}$$
Explanation:
Step 1: Concept used:
We know that the formula of work done- $$ W = FS\ cos\theta$$
if $$\theta = 90^{\circ}$$ then $$W=0$$
                           
Step 2: Conclusion:
The force of gravity on a satellite as it moves around the Earth is perpendicular to its displacement. As a result, the Earth performs no work on the satellite. 
The work done is $$0$$ when the force direction is perpendicular to the displacement.
Hence, the work done by force of gravity on the satellite is $$Zero$$.
$$ W =0$$


A mass of $$10\ kg$$ is at a point $$A$$ on a table. It is moved to a point $$B$$. If the line joining $$A$$ and $$B$$ is horizontal what is the work done on the object by the gravitational force? Explain your answer.

Hint: Observe the angle between force of gravity and displacement.

Explanation:

Step 1: Concept used:
Work done is given by:
$$W= \overrightarrow {F}. \overrightarrow {S}$$
$$W=F\ S\cos\theta$$
where, $$F = $$ force applied
$$S = $$ displacement
$$\theta = $$ angle between the force applied and displacement.

Step 2: Finding the work done:
In the given problem, the force of gravity and displacement are perpendicular to each other. 

$$\theta =$$ $$90$$
$$ \therefore W=0$$ 

Hence, the work done is zero.


Look at the activities listed below. Reason out whether or not work is done in the light of your understanding of the term 'work'.
(A). Suma is swimming in a pond.
(B). A donkey is carrying a load on its back.
(C). A wind-mill is lifting water from a well.
(D). A green plant is carrying out photosynthesis.
(E). An engine is pulling a train.
(F). Food grains are getting dried in the sun.
(G). A sailboat is moving due to wind energy.

Work done            $$W   =  FS$$ $$cos\theta$$    where   $$\theta$$ is the angle between $$F$$ and $$S$$

(A)  :     While swimming, Suma pushes the water in the backward direction and as a result of Newton's 3rd law of motion, the water applies an equal force on Suma in forward direction. Hence Suma swims in the forward direction.
        $$\therefore$$ In this case,   $$\theta  = 0^o$$         $$\implies  W = FS  cos0^o =  FS$$

(B) :      Donkey applies a force in the vertical direction to carry the load and its displacement is in the horizontal direction, thus  $$\theta = 90^o$$  in this case.
       $$\implies  W = FS  cos90^o =  0$$

(C)  :     Wind-mill applies a force in the vertical direction to lift the water from well and the water also moves in the vertical direction.
        $$\therefore$$ In this case,   $$\theta  = 0^o$$        $$\implies  W = FS  cos0^o =  FS$$

(D)   :    During photosynthesis, all the metabolic reactions take place inside the leaf of the plant, but there is no displacement either of the plant or the leaves.
     $$\therefore$$   $$S = 0$$                 $$\implies  W = 0$$

(E)   :      While pulling, the engine applies a force in the forward direction on the train and the train also moves in the forward direction. 
      $$\therefore$$ In this case,   $$\theta  = 0^o$$         $$\implies  W = FS  cos0^o =  FS$$

(F)   :   When the food grains are getting dried in the sun, they remain still at their position i.e the food grains do not get displaced.
     $$\therefore$$   $$S = 0$$                 $$\implies  W = 0$$

(G)   :     Wind  applies a force on the sailboat in the forward direction making the boat to move in the direction of force. 
        $$\therefore$$ In this case,   $$\theta  = 0^o$$         $$\implies  W = FS  cos0^o =  FS$$


The potential energy of a freely falling object decreases progressively. Does this violate the law of conservation of energy? Why?

No, it does not violate the law of conservation of energy.
According to conservation of energy,         $$E_{Total} = P.E+K.E$$
Thus for a freely falling object, its potential energy decreases but kinetic energy increases in such a way that its total mechanical energy remains constant.
Moreover, potential energy of a freely falling object is converted into its kinetic energy.


An object of Mass $$40\ kg$$ is raised to a height of $$5m$$ above the ground what is its potential energy? If the object is allowed to fall, find its kinetic energy when it is half way down.

Given :         
 Mass of the object        $$M = 40  kg$$
 Height upto which it is raised      $$h = 5$$ m
Potential energy        $$P.E = Mgh$$
$$P.E = 40 \times 10 \times 5  =  2000  J$$

Let its kinetic energy be $$E_k$$ when it is half way down   i.e  $$h' = \dfrac{h}{2}$$
According to conservation of energy,           
 $$K.E_i + P.E_i = K.E_f + P.E_f$$
$$0+ 2000  =  E_k + Mg\dfrac{h}{2}$$
$$\therefore$$   $$2000  =  E_k + 40 \times 10 \times \dfrac{5}{2}$$
$$\implies  E_k = 1000  J$$


Does the transfer of energy take place when you push a huge rock with all your might and fail to move it? Where is the energy you spend going?


Explanation: 
Step 1: Concept used: 
The energy transition is the process through which energy transforms from one form to another. Energy conservation refers to the fact that, while energy can be transformed but the total amount of energy does not vary.
Step 2: Conclusion:
There is no muscular energy transfer to the stationary rock when we push a massive rock. Furthermore, there is no energy loss because muscular energy is converted to thermal energy, causing our bodies to heat up. We know that energy can't be created or destroyed; it can only be converted from one form to another because of the conservation of energy principle.


State whether the following objects possess kinetic energy , potential energy or both:
a) a man climbing a hill
b) a flying aeroplane
c) a bird running on the ground
d) a ceiling fan in the off position
e) a stretched spring lying on the ground

a) Both kinetic and potential energy
b) Both kinetic and potential energy
c) Only kinetic energy
d) Only potential energy
e) Only potential energy


In each of the following a force, $$F$$ is acting on an object of mass $$M$$. The direction of displacement is from west to east shown by the longer arrow. Observe the diagram carefully and state whether the work done by the force is negative, positive, or zero.
464569_28160575da51458d945d52df0796aba1.png

Workdone         $$W = \vec{F}. \vec{S}   =F  S  $$ $$cos\theta$$           where  $$\theta$$ is the angle between $$\vec{F}$$  and  $$\vec{S}$$
For (a)  :       $$\theta = 90^\circ$$
             $$\therefore$$  $$W =F  S  $$ $$cos90^\circ  =  0$$    

For (b)  :       $$\theta = 0^\circ$$
             $$\therefore$$  $$W =F  S  $$ $$cos0^\circ  =  FS  > 0$$   

For (c)  :       $$\theta = 180^\circ$$
             $$\therefore$$  $$W =F  S  $$ $$cos180^\circ  = - FS  < 0$$   

490204_464569_ans_724670722be442f88c385b63277f68cb.png


A car at rest has mass $$1000 \ kg.$$ It is moved by a horizontal force of $$50 \ N$$ over a horizontal surface. Calculate the work done by the force in $$8 \ s$$.

If $$ a $$ be the acceleration of the car, 
Then by Newton's second law of motion 
$$F=ma$$
$$\Rightarrow a=\dfrac{F}{m}=\dfrac{50}{1000}=0.05 m/s^2$$

By second equation of motion,
$$S=ut+\dfrac{1}{2}at^2$$
Given:
Time $$t=8 \ s$$
Initial velocity $$u=0$$
The displacement in $$8 \ s$$ is 
$$S=0(8)+\dfrac{1}{2}(0.05)(8)^2=1.6 \ m$$

Work done by the force in $$8 \ s$$ is,
$$W=F \times S=(50 \ N) \times  (1.6 \ m)=80 \ J$$


In which of the following cases is the work done positive or zero or negative?
a) Work done by the porter on a suitcase in lifting it from the platform on to his head.
b) Work done by the force of gravity on suitcase as the suitcase falls from porter's head.
c) Work done by the porter standing on platform with suitcase on his head.
d) Work done by force of gravity on a ball thrown up vertically up into the sky.
e) Work done by force applied by hands of a man swimming in a pond.

1) Positive = Force and displacement in same direction.
2) Positive = Force of gravity and displacement by falling  are in same direction.
3) Zero = No displacement, therefore, no work.
4) Negative = Force and displacement in opposite direction.
5) Negative = Force and displacement in opposite direction.


Define work and write its units.

Work done by a force acting on an object is equal to the magnitude of the force multiplied by the distance moved in the direction of the force. We define work to be equal to the product of the force and the displacement. 
Work done = force × displacement 
W = F × s

$$1J=1N\times 1m$$
$$J=Nm$$ 
where Joules $$(J)$$ is the unit of work.


Unit of energy is ________.

The Joule, symbol J, is the unit of energy in the International System of Units.


Give few examples where displacement of an object is in the direction opposite to the force acting on the object.

1. The force that friction applies on any object is always opposite in direction of the movement of that object.
2. The gravity acting on an object being lifted is always opposite in direction of the movement of that object.


When you lift a box from the floor and put it on an almirah the potential energy of the box increases but there is no change in its kinetic energy. Is it violation of conservation of energy? Explain.

When a person lifts a box from the floor and places it on an almirah, the potential energy increases because the person's force on the box is in the opposite direction and of equal magnitude to the gravitational force acting on the box. The box's velocity will not change as a result of this. There will be no change in kinetic energy as a result.
When no external force is applied, the energy of the system is conserved. The box has been worked on by an external force to enhance its potential energy. It is not a violation of energy conservation.


What is the work done by the moon when it revolves around the earth?

The work done by the moon when it revolves around the earth is zero because here displacement of moon and gravitational force of earth are perpendicular to each other.
i.e., Work done (W) $$= F \times d \times \cos { \theta  } $$
where F = applied force
d = displacement of the body
$$\theta$$ = angle between the applied force and the displacement of the body
In the case of revolution of moon around the earth, $$\theta$$ is $$90°$$. Thus, $$ W = F \times d \times \cos { 90°  } $$
 = $$F \times d \times 0 = 0$$
So, the work done by the moon when it revolves around the earth is zero.


The SI unit of work is__________ (watt, joule).

The SI unit of work is the joule (J), which is defined as the work done by a force of one newton through a displacement of one metre.


Energy can neither be ______ nor _____. It can only be changed from one form to another.

Energy can neither be $$created$$ nor $$destroyed$$. It can only be changed from one form to another.


Work done by the force is said to be negative, if the displacement of a body is_______. (along the force, against the force)

Ans - Against the force
Work done is negative when the force acts opposite to the direction of displacement.


Capacity of doing work is called_____________ (power, energy).

Energy is defined as the capacity to do work


Explain Positive, Negative and Zero Work. Give one example of each.

POSITIVE WORK: The work done on an object is said to be positive work when force and displacement are in same direction.
Example: When an object moves on horizontal surface, force and displacement acts in same direction. So, work done is positive.

NEGATIVE WORK: The work done is said to be negative work when force and displacement are in opposite direction.
Example: When an object is thrown upwards,the force of gravity is in downward direction whereas displacement acts in upward direction. 

ZERO WORK: The work done is said to be zero when force and displacement are perpendicular to each other or when either force or displacement is zero.
Example: When we hold an object and walk, the force acts in downward direction whereas displacement acts in forward direction.



A coolie with a load on his head walking on a horizontal road. What is the work done against gravity?

Work done against gravity is zero as there is no displacement along or opposite to the direction of force of gravity.


Explain by an example that a body may possess energy when it is not in motion.

Every body which is not in motion lack kinetic energy but it can have potential energy.
For example : a rock kept on a roof does not have ant kinetic energy but it has potential energy equal to
                                                                       PE = mgh
                                                        where,  m = mass of the stone
                                                                      h = height at which the stone is kept
    
 


Give the formula for calculating work done. What is the SI unit of work?


We define work to be equal to the product of the force and displacement. 
Work done = force × displacement 
W = F × s
The SI unit of work is Joule (J). 


A force of $$10N$$ moves a body with a constant speed of $$2m{s^{ - 1}}$$. Calculate the power of the body.

Given Force $$F=10 N$$ and velocity $$v=2 m/s$$

Power, $$P=\dfrac{\text{Work done}}{Time}$$

Also we know, work done $$W=F\times s$$ where s= displacement  &  velocity, $$v=\dfrac{s}{t}$$

Thus, power, $$P=\dfrac{F\times s}{t} = F \times v $$

$$P=10\times 2=20\ W$$


A boy mass of 50kg runs up a staircase of 45 steps in 9s.If the height of each step is 15cm.Find his power.Take ($$g={ 10 }{ m }{ s^{ -2 } }$$)


1328245_1115797_ans_cffb1c8c7b0c447aa45d7b1b0f560ef4.jpg


The power of heart which beats 72 times in a minute is 1.2 kW. Calculate the work done by the heart for each heart beat.

$$Power=\dfrac{work \ done}{time}$$
$$P=1200W$$
$$t=\dfrac{60}{72} s$$
$$W=P\times T$$
$$= 12000\times \left(\dfrac{60}{72}\right)$$
$$= 1000J$$


           
An engine pulls a train 1 Km over a level track with constant speed. Calculate the work done by the train. Given that the frictional resistance $$ 5 \times 10^5 N $$.

Given that,

Distance $$d=1\ km=1000\,m$$

Force $$F=5\times {{10}^{5}}\,N$$

$$\because $$  Train is moving with constant speed then,  

Force applied by the engine = Friction force 

Now, the work done by train = work done by friction force

  $$ W=F\centerdot s $$

 $$ W=5\times {{10}^{5}}\times 1000 $$

 $$ W=5\times {{10}^{8}}\,J $$

Hence, the work done by the train $$5\times {{10}^{8}}\,J$$ 



Define energy.

Energy is the capacity to do work. The SI unit of energy is $$Joule\ (J)$$.
The various forms include potential energy, kinetic energy, heat energy, chemical energy, electrical energy and light energy.



Define the following term:
Potential energy.

The energy stored in a body due to its position and by the virtue of its state of strain is called potential energy.

When a body of mass $$m$$ is at a height $$h$$ from the ground, the potential energy of the body is $$U=mgh$$.



Seema tried to push a heavy rock of $$100 kg$$ for $$200s$$ but could not move it. Find the work done by seema at the end of $$200s$$.

Given that,
Seema tried to push a heavy rock of 100kg for 200s but could not move it.

Since the rock does not move 
So displacement of rock is zero 
So, Work done by seema will be zero


If a particle falls down a height, its potential energy decreases. Describe whether it is violating the law of energy.

As the particle is falling down its potential energy ($$U=mgh$$) is decreasing as the height of the particle from the ground is decreasing. But at the same time, the velocity of the particle is increasing as it is falling under the action of gravity. So, the particle has an acceleration $$g$$. Therefore, the kinetic energy ($$T=\dfrac{1}{2}mv^2$$) of the particle is increasing.

$$\text{The loss of potential energy}=\text{The gain of kinetic energy}$$

Therefore, in this process, the conservation of energy is not violated.



Consider a one-dimensional motion of a particle with total energy $$E$$. There are four regions. $$A,B,C$$ and $$D$$ which the relation between potential energy $$V$$, kinetic energy $$(K)$$ and total energy $$E$$ is as given below:
Region $$C:K > E$$

Region $$C: K > E$$
$$E=V+K$$
$$V=E-K$$
$$\Rightarrow \ V < 0\ PE$$ is negative
This is possible because $$PE$$ can be negative.


A rocket of mass $$3\times { 10 }^{ 6 }\ kg$$ takes off from a launching pad and acquires a vertical velocity of $$1 kms^{-1}$$ and an altitude of $$25\ km$$. Calculate its (a) potential energy (b) kinetic energy.

Given;
$$m=3 \times10^6\ kg$$; mass of the rocket
$$ g= 9.8\ ms^{-2}$$; acceleration due to gravity
$$ h =25\ km$$; height of the rocket from ground
$$\Rightarrow h =25000\ m$$
$$ v = 1\ kms^{-1}$$; velocity of the rocket
$$\Rightarrow v = 1000\ ms^{-1} $$

(a) Potential energy of the rocket-
$$P.E. =mgh$$
$$\Rightarrow P.E. = 3\times 10^6 \times 9.8 \times 25000\ J$$
$$\Rightarrow P.E. = 735\times 10^9\ J$$

(b) Kinetic energy of the rocket-
$$K.E=\dfrac{1}{2}mv^2$$
$$\Rightarrow K.E. = \dfrac{1}{2} \times 3 \times 10^{6} \times (1000)^2 J$$
$$\Rightarrow K.E. = 1.5 \times 10^{12} J$$


A body is thrown up with a kinetic energy of 10J. If it attains a maximum height of 5m, find the mass of the body.

Here, kinetic energy of the body, $$ K.E =10\ J$$Ek=10J
Height attained by the body, $$ h = 5\ m$$
h=5m
When the body attains maximum height, its kinetic energy is converted into potential energy.

$$P.E=K.E $$

$$\implies  mgh =10$$

$$\implies m =\dfrac{10}{gh} = \dfrac{10}{10 \times 5} =0.2\ kg$$


A projectile is fired from the top of a $$40\ m$$ high cliff with an initial speed of $$50\ m/s$$ at an unknown angle. Find its speed when it hits the ground.

Given , height $$h = 40 \ m$$
initial speed $$ u = 50 \ m/s$$

Applying law of conservation of energy,

$$mgh+\dfrac{1}{2}mu^2=\dfrac{1}{2}mv^2$$

where, potential energy $$ U = mgh$$
initial kinetic energy $$K_i = \dfrac 12 mu^2$$
final kinetic energy $$K_f = \dfrac 12 mv^2$$

$$\Rightarrow gh+\dfrac{1}{2}u^2=\dfrac{1}{2}v^2$$

$$10 \times 40+\dfrac{1}{2}\times 50^2=\dfrac{1}{2}v^2$$

$$v^2=3300$$

$$\therefore v=57.4\approx 58 m/s$$


Give two examples of (i) positive work (ii) negative work (iii) zero work

Positive work -
1. Pushing a box through a distance.
2. Lift a book through a height.

Negative work -
1. Work done by force of gravity on an object when lifted.
2. Work done by friction when we walk.

Zero work -
1. Pushing a wall.
2. Walking while holding an object.


Suppose a hammer of mass 1 kg is falling freely on the wooden block, if the hammer fall from a height of 1 m. How much kinetic energy just before hitting the wooden block? 

Given,
$$h=1\ m\\m=1\ kg$$

When hammer will reach on the ground, its potential energy will convert into kinetic energy.
From conservation of energy,
Loss in potential energy = Gain in kinetic energy
$$mgh=K.E$$
$$K.E=1\times 10\times 1=10\ N-m\ or\ jul$$



What happens to the work done when the displacement of a body is at right angles to the direction of force acting on it? explain your answer.

If you applied a force at right angles to the motion,

 you would turn the velocity to the direction of the force.

If you have a rock on a string, and give it a few whirls around,

 the applied force (tension of the string) is always at right angles to the motion, and it goes in a circle.



Energy can be defined as:

Energy is the capacity for doing work. It may exist in potential,kinetic, thermal, electrical, chemical, nuclear, or other various forms.


Is work a scalar or a vector quantity ?

Work is dot product of vectors force and displacement , and dot products are scalar. Therefore work is an scalar quantity. 
Another way of telling this is work has only magnitude and no direction, so it is a scalar


A boy weighing $$40\ kg$$ makes a high jump of $$1.5\ m$$ vertically upwards. 
(i) What is his kinetic energy at the highest point?
(ii) What is his potential energy at the highest point?

Mass of the boy, $$m = 40\ kg$$
Height, $$h = 1.5\ m$$
At highest point,
velocity, $$v= 0\ m/s$$
(i) Kinetic energy at the highest point, $$KE=\dfrac 12mv^2$$
$$KE=\dfrac 12\times 40 \times (0)^2$$
$$KE=0\ J$$

(ii) Potential energy at the highest point, $$U=mgh$$  (where $$g = \text{acceleration due to gravity}$$
$$U= 40\times 9.8\times 1.5$$
$$U=588\ J$$


A car of mass $$1200$$ kg is moving with a speed of $$81$$ km/hr. Calculate its kinetic energy.

Given ,
Mass of car, $$m= 1200 \ kg$$
Speed of car, $$v= 81 \ km/hr =81 ( \dfrac{1000}{60 \times 60}) m/s = 22.5\ m/s$$

Kinetic energy , $$K=\dfrac12 mv^2= \dfrac 12 \times 1200 \times 22.5^2=303750\ J$$


A body does 20 J of work in 5 s. What is its power? 

Given,
$$ W = 20\ J $$ 
$$ t = 5\ s$$ 

Power $$ P = \dfrac{W}{t} $$ 

           $$ P = \dfrac{20}{5} = 4\ W $$ 



 1 kW = _______ watts 

$$1000 \ watts$$


The potential energy of your body is least when you are....

The potential energy of your body is least when you are lying down on the ground. Because in this stage, your maximum body parts come in contact with earth's surface, its mean that, effective Height of body weight becomes lower as possible. Hence potential energy of found minimum in this stage.


Define 1 joule of energy. 

The joule (J) is the SI unit of energy.

Energy is a measure of the capacity to do work. One joule equals the work done (or energy expended) by a force of one newton (N) acting over a distance of one meter (m).



When the work is said to be done by a force?

Work is said to be done by the force when direction of force and motion of object is same.


Illustrate the law of conservation of energy by discussing the free fall of a body.

According to law of conservation of energy, energy can neither be created nor be destroyed. It can only be transformed from one form to another. whenever energy changes from one form to another, the total amount of energy remains constant.
Law of conservation of energy is valid in all situations and for all kinds of energy transformations.
For example: When a body is at some height from earth it possess only potential energy, as the body falls downwards then its potential energy goes on decreasing but kinetic energy goes on increasing.
Just before hitting the ground entire potential energy is transformed into kinetic energy and it possesses only kinetic energy.
Total energy = kinetic + potential energy
At maximum height, kinetic energy = 0
Total energy = potential energy
On hitting ground, potential energy = 0
Total energy = kinetic energy


Differentiate between work and power.

 Work Power
 Work is said to be done, when a force causes displacement. Power is defined as the work done per unit time or rate at which energy is consumed.
 SI unit of work is Joule. SI unit of power is Watt.
 Work done $$W = force \times displacement$$(in the direction of force). Power $$P = \dfrac{Work}{time}$$
 Work done is independent of time taken. Power of an object/body depends on the time taken,



State the SI unit of work and power.

The SI unit of work is $$Joule\, (J)$$. It is defined by the work done by $$1\, N$$ force for a displacement of $$1\, m$$.
The SI unit of power is $$Watt\, (W)$$. It is defined as $$1\, J$$ work done in $$1\, s$$.


Compare the power at which each of the following is moving upwards against the force of gravity?  $$(given\ g=10\ ms^{-2})$$
a butterfly of mass $$1.0\ g$$ that flies upward at a rate of $$0.5\ ms^{-1}$$.

$$power = mg \times velocity,\ g=10ms^{-2}$$
$$=\dfrac{1}{1000} \times  10 \times 0.5\ W$$
$$=\dfrac{0.5}{100} = 5 \times 10^{-3 } \ W$$


A boy of mass $$50 \,kg$$ runs up a staircase of $$45$$ steps in $$9s$$. If the height of each step is $$15 \,cm$$, find the power consumed by him.
Take, $$ \,g = 10 \, m/s^{-2}$$

Given,
mass of boy $$m = 50\, kg$$,
time taken $$t = 9\, s$$,
height of one step $$h = 15\, cm$$,
acceleration due to gravity $$g = 10 m/s^2$$.
Now.
weight $$mg= 50\times 10 = 500N$$
$$Work=Potential$$ $$Energy$$   (Because work done by him is converted into gain in potential energy)
Total height gained=$$H$$=total height of staircase = $$45\times\dfrac{15}{100}=6.75 \,m$$
$$Power=\dfrac{Work}{Time}=\dfrac{mgH}{t}$$

$$\Rightarrow P = \dfrac{500\times6.75}{9}=\dfrac{3375}{9} W$$
$$ Power\, consumed = 375\, Watt$$


A girl having mass of $$35\ kg$$ sits on a trolley of mass $$5\ kg$$. The trolley is given an initial velocity of $$4\ ms^{1}$$ by applying a force. The trolley comes to rest after traversing a distance of $$16\ m$$. How much work is done on the trolley ? 

Given:
Mass of girl $$m_g=35\ kg$$ 
Mass of trolley $$m_t=5kg$$
Initial velocity $$\ u=4\ m/s$$ 
FInal velocity $$\ v=0$$ and distance $$s=16m$$

Work done $$W= F \times s$$
By Newton's second law of motion, 
$$F=m_oa$$ 
where $$m_o=m_g+m_t$$ is the mass of girl and trolley together.
$$m_o=35+5=40 \ kg$$

So $$W=(m_o a) \times s$$
We need to calculate acceleration.
By third equation of motion,
$$v^2-u^2=2as$$
$$\Rightarrow a=\dfrac{\nu ^2 -u^2}{2s}=\dfrac{0-16}{2 \times 16}=-0.5\ ms^{-2}$$

Using the above value of $$a$$, work done on trolley can be calculated as follows:
$$W=(m_oa) \times s=-(40 \times 0.5) \times 16=-320\ J$$
The negative sign indicates that the force of trolley and its displacement are in opposite direction.


Compare the power at which each of the following is moving upwards against the force of gravity?  $$(given\ g=10\ ms^{-2})$$
a $$250-g$$ squirrel climbing up on a tree at a rate of $$0.5\ ms^{-1}$$.

The power is given as:
$$power =\dfrac{250}{1000} \times  10 \times 0.5\ W$$

$$=\dfrac{1}{4} \times 10 \times 0.5 = 1.25 \ W$$


A body is moved along a closed loop. Is the work done in moving the body necessarily zero? If not, state the condition under which work done over a closed path is always zero.

Work done by a body moving along closed loop can be zero if only conservative force acting on the body during motion.
Work done by a body moving along a loop is not zero if any non-conservative force, i.e., frictional, electrostatic, magnetic force are acting on body.


Consider a one-dimensional motion of a particle with total energy $$E$$. There are four regions. $$A,B,C$$ and $$D$$ which the relation between potential energy $$V$$, kinetic energy $$(K)$$ and total energy $$E$$ is as given below:
Region $$A:V > E$$

For region $$A:V > E$$
$$E=V+K$$
$$K=E-V$$
$$\therefore \ V > E$$. So, $$K < 0$$ or $$KE$$ is negative. Which is not possible.


Calculate the power of a crane in watts, which lifts a mass of $$100\ kg$$ to a height of $$10\ m$$ in $$20$$ seconds.

$$ m=100\ kg \quad g =10\ m/s\quad h=10\ m$$
$$W = mgh= 100 \times 10 \times 10 = 10000\ J$$

$$P=\dfrac{W}{t}= \dfrac{10000}{20}=500\ W$$


The average work done by a human heart while it beats once is $$0.5\ J$$. Calculate the power used by heart if it beats $$72$$ times in a minute.

$$P=\dfrac{W.D.}{time}$$

$$WD$$ in $$1$$ beat by heart $$=0.5\ J$$
$$WD$$ in $$72$$ beats by heart $$=0.5\times 72=36\ J$$
Time $$=1$$ minute $$=60\ s$$

$$\therefore P=\dfrac{W.D.}{t}=\dfrac{36}{60}=\dfrac{6}{10}=0.6\ W$$

Hence, power used by heart to beat 72 times a minute is $$0.6\ W$$


The energy of a body is its capacity to do _____.

The energy of a body is its capacity to do $$work$$.


A man spends $$6.4\,KJ$$ energy in displacing a body by $$64 \, m$$ in the direction in which he applies force, in $$2.5\, s$$ Calculate:
(i) the force applied and
(ii) the power Spent (in HP) by the man

$$\textbf{Step 1: Force applied }$$
$$(i)$$ Work done by the man $$=$$ Force $$\times$$ distance moved in direction of force applied. 
$$W=F \times S$$
$$\Rightarrow 6.4 \times  10^3\ J=F \times 64m$$ 
$$\Rightarrow F =\dfrac {6.4\times 10^3}{64}= 100\, N$$

$$\textbf{Step 2: Power spent}$$
 $$(ii)$$ Power spent $$ =\dfrac{\text{Work done}}{\text{Time}}=\dfrac {6.4\times 10^3\ J}{2.5\ s}= 2560\, W$$

$$[\because 1\, H.P= 746\, W  \Rightarrow 1\ W=\dfrac{1}{746}H.P.]$$

$$\therefore $$ Power spent in HP = $$ 2560\, W =\dfrac {2560}{746} H.P = 3.43 \, H.P$$


A ball of mass $$0.20
\, kg$$ is thrown vertically upwards with an initial velocity of $$20 \, m\, s^{-1}$$. Calculate the maximum potential energy it gains as it goes up.

From conservation of energy:
Gain in potential energy = Lass in kinetic energy

                       $$\Delta U = \dfrac{1}{2} mv^2$$

                       $$\Delta U  = \dfrac{1}{2}\times  0.20 \times 20 \times 20 = 40\,J$$ 


A body of mass $$5\, kg$$ falls from a height of $$10\, m$$ to $$4 \,m$$. Calculate: (i) the loss in potential energy of the body. (ii) the total energy possessed by the body at any instant? (Take $$g = 10\, m \,s^{-2}$$)

(i) Mass of the body $$= 5 \,kg$$  $$g=10m/s^2$$   
P.E. at height $$10\,m=  mgh = 5 \times 10 \times 10 = 500\,J$$ 
P.E. at height $$4 \,m = mgh = 5 \times 10 \times 4 = 200 \, J$$ 
Loss in P.E. $$= (500 - 200)\, J= 300\,J$$ 
(ii) The total energy possessed by the body at 
Any instant remains constant for free fall 
It is equal to the sum of P.E. and K.E. 
$$\therefore$$ At height $$10\, m$$,  i.e.. at top most point. $$K.E. = 0$$ 
$$\therefore$$ Total energy $$= (PE_{Top}+ KE_{Top}) $$
Total energy $$= 500 + 0 = 500\, J$$ 


Write details answer?
Explain the difference between potential energy and kinetic energy.

Potential Energy   Kinetic Energy 

It is possessed by a body virtue of configuration and position.		  It is possessed by a body virtue of its motion.
eg: A book kept on a shelf has potential energy.eg: A moving car posses kinetic energy


What do you mean by potential energy? Explain with example.

The energy which a body possesses due to its position or due to change in the shape is called potential energy.

Example 1: A brick that is suspended above the ground has energy because it would do work by falling to the ground. Its energy is due to its position and therefore it is potential energy.

Example 2: The potential energy of a stretched bow string results from a change in its shape.


What is necessary for work to be done on a body?

For work to be done on a body, action of force and displacement in the direction of the force is necessary.


Give one example where work done on an object is zero.

A person is walking horizontal straight road with constant speed with a load on his head. In such case, no work is done on the load. This is because the direction of force acting on the load and the displacement seen is perpendicular. 
$$W = 0$$ 

1751737_1832205_ans_8fc95b1570a2426384d9422e32ef513d.png


A man does 60 J of work in 6 seconds. Calculate the power.

$$Power = \dfrac{Work}{time}=\dfrac{60 J}{6 s}$$

$$\therefore  Power=10 watts.$$


Give two situations where energy is supplied but no work is done.

Work done is defined as, $$W=Fs\cos\theta$$
where $$F$$ is the force applied, $$s$$ is the displacement and $$\theta$$ is the angle between the force applied and displacement.

If $$s=0$$ then $$W=0$$
For example,

(i) A person pushing a heavy rock is using energy but if the rock does not move no work is done. 
(ii) A person standing with heavy load on his head. Energy is used/spent in doing this but no work is done, since there is no displacement.


State carefully if the following quantities are 'positive' or 'negative':
Work done by an applied force on a body moving on a rough horizontal plane with uniform velocity.

Here the body is moving on a rough horizontal plane. Frictional force opposes the motion of the body. Therefore, in order to maintain a uniform velocity, a uniform force must be applied to the body. Since the applied force acts in the direction of motion of the body, the work done is positive.



State carefully if the following quantities are 'positive' or 'negative':
Work done by the resistive force of air on a vibrating pendulum in bringing it to rest.

'Negative'.
Reason: The resistive force opposes the motion $$(\theta =0^{0})$$..


Define one Joule.

Joule is the unit of work or energy. Work done is said to be $$1$$ joule, if a force of $$1$$ newton displaces a body through $$1$$ meter in the direction of force.


When is work said to be done on a body? Explain.

Work is said to be done when the force applied on a body makes the body move (i.e., there is the displacement of the body).
For example, a man pushing a car and a cyclist while pedaling a cycle does work, but a man who tries to push a wall does not do any work as the wall does not move.


Give example for the kinetic energy.

The kinetic energy of a body is by the virtue of its motion. Thus, an moving object will possess kinetic energy.
The moving air , the moving ball, the flowing water, the moving striker have kinetic energy.


A body of mass $$10 kg$$ is kept at a height of $$5 m$$. It is allowed to fall and reach the ground.
What is the kinetic energy possessed by the body just before hitting the ground? Take $$g= 10 m s^{-2}.$$

$$m=10kg$$   $$g=10m/s^2$$   $$h=5m$$
Initial energy of the system=$$P.E = mgh=$$$$10\times10\times5=500J$$
Since total mechanical energy (KE+PE) remains conserved. 
At ground height is zero so PE become zero ,all PE is lost and converted into KE.
Therefore, initial potential energy = Final kinetic energy 
So in this case when body reachs the ground potential energy would be equal to zero.
So, final kinetic energy = initial potential energy = $$500J$$


If a man of mass 50 kg takes 60 s to climb up 20 steps, each 15 cm high, calculate his power.

$$m = 50 kg , g = 10 m/s^{2}$$
$$h=15cm\times 20=300cm =3m$$
$$t=60s$$
$$work, W= mgh = 50\times 10\times 3=1500J$$
$$power P =\dfrac{w}{t}=\dfrac{1500}{60}=25W$$


State the principle of conservation of energy.

The law of conservation of energy states that the total energy of an isolated system remains constant, it is said to be conserved over time. This law means that energy can neither be created nor destroyed; rather, it can only be transformed from one form to another.


How is work measured?

The work is calculated by multiplying the force (F) by the amount of movement (d) of an object. 
W = F * d 
Ex:- A force of 10 newtons move an object 3 meters, does 30 N-m of work. 
A Newton-meter is the same thing as a joule.


Look at the activities listed below. Reason out whether work is done or not done in the light of your understanding of the term 'work'.
(a) Suma is swimming in a pond.
(b) A donkey is carrying a load on its back.
(c) A windmill is lifting water from a well.
(d) A green plant is carrying out photosynthesis.
(e) An engine is pulling a train.
(f) Food grains are getting dried in the sun.
(g) A sailboat is moving due to wind energy.

(a)Work is done because the displacement of swimmer takes place and a force is applied.
(b)Donkey is not doing any work in carrying the load as weight (acting vertically downward) and displacement (acting horizontal) are perpendicular to each other. But, it does work against friction between its hoofs and the ground.
(c) Work is done because the displacement of water takes place and a force is applied by the wind mill. Wind mill does a work against gravity.
(d) No work is done, because no displacement takes place.
(e) Work is done because the displacement of train takes place and a force is applied by the engine.
(f) No work is done because displacement does net take place.
(g) Work ts done because the displacement of sailboat takes place and a force is applied by the wind.


What is the amount of work done by:
(a) A weight - lifter in holding a weight of $$120\ kg$$ on his shoulder for $$30\ s$$.
(b) A locomotive against gravity, if it is travelling on a plane road.

Work done = Force x displacement

(a) There is no displacement of weight-lifter. Hence, no work done by weight-lifter.
(b) Work done by locomotive against gravity is zero, as it has no displacement along or opposite to the direction of gravity.


Which person did the maximum work?
a) Venu walks 1,000 meters to the store. He buys 4.448 N of chocolates and then carries it to his friend's house which is 500 meters away.
b) Kala lifts her 22 N cat to a distance of 0.5 meters.
c) Ramu carries groceries from a car to his house. Each bag of groceries weighs 40 N. He has 10 bags. He lifts each bag up 1 meter to carry it and then walks 10 m from his car to his house.

The work done by force is given as,
$$\text{Work} = \text{Force} \times \text{Displacement in the direction of force}$$

(a) No work is done by venu since the force of gravity acting on chocolates is perpendicular to their displacement.

(b) $$W= F \times s $$
$$\Rightarrow W = mgh=22\times 0.5$$
$$\Rightarrow W = 11\ N$$

(c) $$W=10 \times mgh$$
$$\Rightarrow W=10\times 40\times 1$$
$$\Rightarrow W=400\ N$$

Therefore, Ramu did maximum work.


A man moves on a straight horizontal road with a block of mass $$2\ kg$$ in his hand. If he covers a distance of $$40\ m$$ with an acceleration of $$0.5\ m/s^{2}$$, find the work done by the man on the block during the motion.

Given, Mass of the block$$=2kg$$

Acceleration$$=0.5m/s^2$$

Force$$=ma=2×0.5N=1N$$

Distance$$=40m$$

Workdone$$=F.S=1\times 40J=40J$$


A  man raises a box of mass $$50 kg$$ to a height of $$2m$$ in $$10 s$$ , while another man rises the same box to the same height in $$ 50 s$$. Compare :
(i) the work done, and 
(ii) the power developed by them.

The work done by each person is,

$$W = 50{\rm{kg}} \times 10{\rm{m}}/{{\rm{s}}^2} \times 2{\rm{m}}$$

$$ = 1000{\rm{\ J}}$$

Power is equal to work done per unit time

Power of first person will be,

$$ = \dfrac{{1000{\rm{J}}}}{{10\sec }}$$

$$ = 100{\rm{W}}$$

Power of second person will be

$$ = \dfrac{{1000{\rm{J}}}}{{50\sec }}$$

$$=20{\rm{W}}$$



What is potential energy ? Mention the factors that the gravitational potential energy of a body depends on.

The energy which a body possess either due to its position or due to its deformation is called potential energy. 
$$Gravitational$$   $$PE = m \times g \times h $$  where $$m$$ is mass of body ,$$g$$ is acceleration due to gravity ,$$h$$ is the height.
Gravitational Potential energy of a body depends on 
  1. the mass of a body and 
  2. the height to which it is raised from ground.


State whether the energy possessed in moving an object is:

$$\textbf{It possess kinetic energy.}$$
As the moving object has a velocity $$v$$ and mass $$m$$
So the $$K.E=\dfrac{1}{2}mv^2$$


A boy takes $$3$$ minutes to lift  $$20 $$ litres of water from a $$20 m$$ deep well, while his father does the same in $$2 $$ minutes. Compare :
(i) the work, and 
(ii) power developed by them

As $$1\;{\rm{litre}} = 1\;{\rm{kg}}$$ therefore, mass of $$20\;{\rm{litre}}$$ water is $$20\;{\rm{kg}}$$.

Both boy and his father does the same work.

Work done by each$$ = mgh = 20 \times 10 \times 20 = 4000\;{\rm{J}}$$

The power is given as,

Power$$ = \dfrac{Work}{Time}$$

Power of boy$$ = \dfrac{{4000}}{{3 \times 60}} = 22.22\;{\rm{W}}$$

Power of father$$ = \dfrac{{4000}}{{2 \times 60}} = 33.33\;{\rm{W}}$$

Thus power of father is one and half times that of boy.



Define the SI unit of power. A lamp consumes 1200 J of electrical energy in a minute. What is its power?

SI unit of power is Watt (W)
$$ P = \dfrac{E}{t}=\dfrac{1200}{1 \times 60}=20\ W$$


(a) State and define SI unit of power.
(b) A person carrying 10 bricks each of mass $$2.5\,kg$$ on his head moves to a height $$20\,m$$ in  $$50\,s$$. Calculate power consumed in carrying bricks by the person.
Take, $$g = 10 m/s^2$$

(a) The rate of doing work is called as the power. The SI unit of power is $$Watt\,(W)$$. 1 Watt is the power of an agent that consumes energy at the rate of $$1\,J/s$$.
(b) Given,
mass of brick $$m = 2.5\,kg$$
$$\Rightarrow weight = mg = 2.5\times 10 = 25 N$$
weight of 10 bricks $$=25\times 10 = 250 N$$
height $$h = 20\,m$$
time $$t= 50\,s$$

$$Power=\dfrac{Work \,done}{Time \,taken}$$

$$\Rightarrow Power=\dfrac{mgh}{t}=\dfrac{250\times20}{50}$$ $$=\dfrac{5000}{50}=100 \,W$$ 

Power consumed in carrying bricks is $$100\,Watt$$.


A coolie is moving on a road with a luggage on his head. Does he perform work against the force of gravity? Give reason for your answer.

A coolie is carrying a luggage on his head moving on ground does no work against the force of gravity as displacement is normal to the direction of force of gravity.


What is power? How you differentiate kilowatt from kilowatt hour? The Jog Falls in Karnataka state are nearly $$20\ m$$ high. $$2000$$ tonnes of water falls from it in a minute.Calculate the equivalent power if all this energy can be utilized? $$(g=10\ ms^{-2})$$

Power is the rate of doing work. Kilowatt is the unit of power and kilowatt hour is the unit of energy.
$$h=20\ m$$ and $$mass=2000 \times 10^3\ kg=2 \times 10^6\ kg$$
$$Power=\dfrac{mgh}{t} =\dfrac{2 \times 10^5 \times 10 \times 20}{60}W$$
$$=\dfrac{4}{6} \times 10^7\ W=\dfrac{2}{3} \times 10^7\ W$$


A raindrop of mass $$1.00\ g$$ falling from a height of $$1\ km$$ hits the ground with a speed of $$50\ m/s$$. Calculate the loss of potential energy($$P.E.$$) and gain in kinetic energy$$(K.E.).$$
Is the gain in $$K.E.$$ equal to loss of $$P.E.$$? $$(g= 10\ m/s^2)$$

Given : Mass of raindrop, $$m=1\ g =10^{-3}\ kg,$$ 
Height, $$ h=1\ km=10^3\ m$$ 
$$g=10\ ms^{-2}$$  
Speed, $$v=50\ ms^{-1}$$

Loss of P.E. of the drop $$=mgh= 10^{-3}\times 10\times10^3=10 J$$

Gain in K.E. of the drop $$=\dfrac{1}{2}mv^2=  \dfrac{1}{2} \times 10^{-3} \times (50)^2 =1.25 J$$

$$ \implies $$ Both are not equal.


Give one example to show that the sum of potential energy and kinetic energy remains constant if friction is ignored.

During the vertical fall of ball, it friction due to air is neglected, the total sum of potential energy and kinetic energy at each point of its path remains same.


A boy of mass $$40$$ kg climbs a staircase of $$30$$ steps, each of $$0.2\ m$$ height, in $$30\ s$$. Calculate power ($$g = 9.8 m/s^2$$) 

Given,
$$Mass(m)=40\,kg$$
$$Height(h)=30 \times 0.2=6\ m$$
$$Time(t)=30\,s$$
We know that,

$$power=\dfrac{\text{Work Done}}{\text{Time Taken}}$$

$$\Rightarrow power=\dfrac{mgh}{t}$$

$$\Rightarrow power=\dfrac{40 \times 9.8 \times 6}{30}=78.4\,W$$


How much energy is gained by a box of mass $$20\, kg$$ when a man
(a) carrying the box waits for $$5$$ minutes for a bus?
(b) nuts canying the box with a speed of $$3\, m \, s^{-1}$$ to catch the bus?
(c) Raises the box by $$0.5\, m$$ in order to place it inside the bus? $$(g = 10\, m\, s^{-2})$$

Mass of box $$= 20\, kg$$ 
(a) Zero work is done as there is no displacement of the man. 
(b) Work done, Kinetic energy of man 
$$= \dfrac{1}{2} \times mass \times (velocity)^2$$ 
$$=\dfrac{1}{2}\times 20\times (3)^2$$
$$=\dfrac{1}{2}\times 20\times 9$$
$$=90J$$
(c) Week done in raising the box, Potential energy $$= mgh$$
$$ U=20 \times 10 \times 0.5 =100\, J$$ 


Define work. Give its S.I unit. Write an expression for positive work done.

Work is said to be done when the force applied on an object produces a displacement of the object in the direction of force applied.
Work = Force x displacement (expression for positive work)
Work done is said to be positive if displacement of the object is along the direction of force applied.S.I unit of work done = Joule.


Find the height from which body of a mass $$50kg$$ should fall in order to have the K.E of a car of mass $$200kg$$ traveling at $$10ms^{-1}$$ Given $$g = 9.8ms^{-2}$$.

Mass of a body, $$m_{1} = 50 kg$$
Mass of car $$m_{2} = 200kg$$
velocity of the car $$v = 10ms^{-1}, g = 9.8ms^{-2}$$
P.E of the body, P.E = $$mgh$$
$$= 50 * 9.8 * h$$
$$= 490h$$.
K.E of the car K.E = $$\frac{1}{2}m_{2} v_{2}^{2}$$
$$=\frac{1}{2} * 200 * 10^{2}$$
$$= 10,000J$$
But P.E of body = K.E of car
$$490h = 10,000$$
$$h =\frac{10,000}{490} = 20.41 m$$.


A $$820\ N$$ Marine in basic training climbs a $$12\ m$$ vertical rope at a constant speed in $$8\ s$$ What is this power output?

Given:
Weight of the marine $$F_g=820 \ N$$
Height climbed $$h=12 \ m$$
Time taken $$t=8s$$

The marine will spend his muscular energy to do work against gravity.
The work done $$W=force \ \times displacement $$
$$W=F_g \times h$$
$$\Rightarrow W = 820 N \times 12m=9840 \ J$$

The displacement of the marine and force applied by him, are both upwards. So the work done is positive.

$$Power=\dfrac{work}{time}=\dfrac{9840}{8}=1230 \ W$$

The power output is $$1230 \ W.$$


Discuss whether any work is being done by each of the following agents and, if so, whether the work is positive or negative.
(a) a chicken scratching the ground
(b) a person studying
(c) a crane lifting a bucket of concrete
(d) the gravitational force on the bucket in part (c)
(e) the leg muscles of a person in the act of sitting down.

(a) Positive work is done by the chicken on the dirt.  
(b) The person does no work on anything in the environment. Perhaps some extra chemical energy goes through being energy transmitted electrically and is converted into internal energy in his brain; but it would be very hard to quantify “extra.”  
(c) Positive work is done on the bucket.  
(d) Negative work is done on the bucket.  
(e) Negative work is done on the person’s torso.


Class 9 Physics Extra Questions