MCQ Questions for Class 9 Physics Work And Energy Quiz 12

A particle is acted upon by a force F which varies with position x as shown in the figure. If the particle at

$x=0$ has the kinetic energy of

$25$ J, then the kinetic energy of the particle at

$x=16$ m is?

0%
$45$ J
0%
$30$ J
0%
$70$ J
0%
$135$ J

An object of mass

$5 \ kg$ falls from rest through a vertical distance of

$20 \ m$ and reaches a velocity of

$10 \ m/s$ . How much work is done by the push of the air on the object?

0%
$-740 \ J$
0%
$-750 \ J$
0%
$-760 \ J$
0%
$-770 \ J$

Explanation

From work energy theorem,

work done by weight + work done by air $=ΔKE$ (change in Kinetic Energy)

${ W }_{ air }=ΔKE−{ W }_{ (weight) }$

$=\dfrac{1}{2}m{ v }^{ 2 }−mgh$

$=\dfrac{1}{2}×5×{ (10) }^{ 2 }−5×10×20$

$=−750J$

An elevator is run by the cables at constant speed. The total work done by the elevator is:

0%
Negative
0%
Positive
0%
Zero
0%
Positive or negative depending on the direction of motion

Explanation

Work done is defined as

$W = Fd \cos\theta$

where,

$F$ is the force applied,

$d$ is the displacement and

$\theta$ is the angle between force applied and displacement.

Since, the elevator is moving at a constant speed, the work done is only due to the force of gravity and the tension force exerted on the elevator by the cable which pulls it up. Since, the force of tension and displacement are in the same direction, the angle

$\theta = 0$ and hence the work done is Positive. Hence, option (B) is correct.

Kinetic energy of the liquid per unit mass is

0%
$\frac { 1 } { 2 } m v ^ { 2 }$
0%
$\frac { 1 } { 2 } v ^ { 2 }$
0%
$\frac { 1 } { 2 } m ^ { 2 } v$
0%
$m v ^ { 2 }$

Explanation

Let

$m$ be the mass of liquid and

$v$ be the velocity of liquid

Hence ,

Kinetic energy of liquid

$,K=\dfrac12 mv^2$

$\therefore$ Kinetic energy of liquid per unit mass

$= \dfrac Km = \dfrac12 v^2$

When the mass of body is halved and velocity is doubled, then the kinetic energy of the body

0%
remains same
0%
is doubled
0%
is 4 times
0%
is $\frac{1}{4}$ th

Explanation

$K.E = \dfrac{1}{2}mv^2$

$New\ K.E = \dfrac{1}{2} (\dfrac{m}{2})\ (2v)^2$ =

$2 \times \dfrac{1}{2}mv^2$ =

$2 \times K.E$

Hence, kinetic energy of body is doubled.

Mass of

$B$ is four times that of

$A . B$ moves with a velocity half that of

$A$ . Then,

$B$ has

0%
kinetic energy equal to that of $A$
0%
half the kinetic energy of $A$
0%
twice the kinetic energy of $A$
0%
kinetic energy one-fourth of $A$

Explanation

According to question given,

$M_B = 4 \times M_A$ and

$V_B = \dfrac{V_A}{2}$

$K.E_A = \dfrac{1}{2}M_AV_A^2$ .........(1)

$K.E_B = \dfrac{1}{2}M_BV_B^2$ =

$\dfrac{1}{2}\ (4 \times M_A) \left(\dfrac{V_A}{2}\right)^2=\dfrac{1}{2}M_AV_A^2$ ........(2)

From above equation

$K.E_A=K.E_B$

A body of mass

$15\ kg$ moving with a velocity of

$10\ ms^{-1}$ is bought to rest. The work done by the brake is

0%
$-250\ J$
0%
$-500\ J$
0%
$-750\ J$
0%
$-1000\ J$

Explanation

$W = change\ in\ kinetic\ energy =\dfrac{1}{2}m(v^2-u^2)$

$W = \dfrac{1}{2} \times 15 \times (0^2-10^2)$

$W = - 750J$

Hence, option

$C$ is correct answer.

If the unit of force and length each be increased by four times, then the unit of energy is increased by

0%
$16$ times
0%
$8$ times
0%
$2$ times
0%
$4$ times

Explanation

Work = Force

$\times$ Displacement ( length)

$\Rightarrow W=FS$

Now,

$F'=4F$

$S' = 4s$

So energy,

$W' = 4F \times 4S$

$\Rightarrow W' = 16F S$

$\Rightarrow W' = 16W$

If the unit of force and length be increased by four times then the unit of energy will increase by

$16$ times.

Work is always done on a body when

0%
a force acts on body to displace it.
0%
it moves through a certain distance.
0%
it experiences an increase in energy through a mechanical influence.
0%
there is no displacement.

Explanation

Work done is given by

$W=Fd$ where

$F$ is Force and

$d$ is displacement

Hence two conditions need to be satisfied for work to be done:

(i) a force should act on an object

(ii) the object must be displaced.

So option A is correct.

The amount of work done on an object when a constant force of

$\dfrac { 1 }{ 10 }\ N$ displaces it by

$10\ cm$ along the line of action of the force is:

0%
$\dfrac { 1 }{ 10 }\ J$
0%
$\dfrac { 1 }{ 100 }\ J$
0%
$1\ J$
0%
$10\ J$

Explanation

Given,

Force applied,

$F= \dfrac {1}{10}\ N$

Displacement,

$d=10\ cm$

$d=\dfrac {10}{100}=\dfrac{1}{10} \ m$

Work done,

$W=F\times d$

$W=\dfrac {1}{10} \times \dfrac{1}{10}$

$W=\dfrac {1}{100}\ J$

A lawn roller is pulled along a horizontal surface through a distance of

$20 \ m$ by a rope with a force of

$200 \ N$ . If the rope makes an angle of

$60^o$ with the vertical while pulling, the amount of work done by the pulling force is:

0%
$3464 \mathrm { J }$
0%
$1000 \mathrm { J }$
0%
$2000 \sqrt { 3 } \mathrm { J }$
0%
$2000 \mathrm { J }$

Explanation

Correct Answer: Option A

Hint: Use the equation for work done, $W=Fscos\theta$ , where F is force, s is displacement and $\theta$ is the angle between force and displacement.

Explanation of Correct Option:

Given, force $F=200N$ , distance $s=20m$ ,
Angle rope make with vertical is $\theta=60^{\circ}$ , so angle rope makes with horizontal $\theta=30^{\circ}$

Work done by pulling force, $W=F.s$

$W=Fscos\theta$

$W = 200 \times 20 \times \dfrac{\sqrt3}{2}$ $\left( {\because \theta = 30^\circ ,\cos 30^\circ = \dfrac{\sqrt3}{2}} \right)$

$W = 3464J$

The amount of work done by pulling force is $3464J$ .

2150669_1320952_ans_9aaa0db4129448dd930bad1822af4c3f.png

The SI unit of work is newton.

0%
True
0%
False

$10 s$ , a body of

$6 kg$ mass is dragged

$8 m$ with uniform velocity across a floor by a steady force of

$20 N$ . The K.E of the body is:

0%
$2.92 J$
0%
$0.92 J$
0%
$1.92 J$
0%
$0J$

The

$SI$ unit of energy, the Joule is equivalent to

0%
$N-m$
0%
$W/s$
0%
$N-m^{2}/s^{2}$
0%
$N/m$

A body of mass m is rotating in a vertical circle of radius 'r' with critical speed. The difference in its K.E. at the top and the bottom is _____.

0%
2mgr
0%
4 mgr
0%
6 mgr
0%
3 mgr

A body of mass 0.5 kg is moving at a constant velocity, work done is

0%
0 J
0%
1.0 J
0%
2.0 J
0%
4.0 J

Explanation

$Work=Force \times displacement$

Since moving with same velocity so no acceleration hence no force . So

$W= 0\times displacement$

$W=0$

In which of the following, law of conservation of energy is violated?

0%
Fusion of nuclei
0%
Mud thrown on a wall
0%
A satellite moving in elliptical orbit
0%
None of these

If the kinetic Energy possessed by a man of

$50$ kg is

$625J$ , the speed of man is

0%
$5m{s^{ - 1}}$
0%
$50m{s^{ - 1}}$
0%
$0.5m{s^{ - 1}}$
0%
$500m{s^{ - 1}}$

A man of mass 20 kg is running in a straight line with velocity 10 m/s. What is value of his kinetic energy?

0%
$10\ J$
0%
$100\ J$
0%
$1000\ J$
0%
$10000\ J$

Explanation

The kinetic energy of the particle is given as-

$K.E. = \dfrac{1}{2}mv^2$

$m$ ; mass of the object

$v$ ; speed of the object

$\Rightarrow K.E. = \dfrac{1}{2} \times 20 \times 10^2J$

$\Rightarrow K.E. = 1000 J$

The power exerted on the body at

$2\ s$ is:

0%
$50\ W$
0%
$100\ W$
0%
$150\ W$
0%
$200\ W$

Two men

$A$ and

$B$ each weighing

$80\ kg$ , run up the same stair case. While

$A$ does it in

$20\ s, B$ does it in

$15\ s$ . The ratio of the powers developed by

$A$ and

$B$ is

0%
$\dfrac{4}{3}$
0%
$\dfrac{3}{4}$
0%
$\dfrac{1}{2}$
0%
$\dfrac{1}{4}$

Explanation

We know

$Power=\dfrac{work }{time}$

work done by both the men is the same = Gain in potential energy =

$m\times g\times h$

$m,g,h$ are same for both so work same for both.

$P \propto \dfrac{1}{t}$

Therefore

$\dfrac{P_{a}}{P_{b}}=\dfrac{t_{b}}{t_{a}} \implies\dfrac{P_{a}}{P_{b}}=\dfrac{15}{20}$ =

$\dfrac{3}{4}$

Hence Option B is correct

Two bodies traveling with velocity

$20ms^{-1}$ and

$40ms^{-1}$ have same K.E. If the mass of the body having velocity

$40ms^{-1}$ is 8 kg the mass of other body is

0%
8 Kg
0%
16 kg
0%
24 kg
0%
32 kg

Explanation

Given

$V_1 = 20 m/s$ and

$V_2 = 40 m/s$

$M_2 = 8\ kg$

$K.E_1 = K.E_2$

$\dfrac{1}{2}M_1V_1^2 = \dfrac{1}{2}M_2V_2^2$

$\implies M_1 = M_2 \times \dfrac{V_2^2}{V_1^2} = 8\times \dfrac{40^2}{20^2} = 32\ kg$

Which of the following does not possess the ability to do work because of motion?

0%
A sparrow flying in the sky
0%
A sparrow moving slowly on the ground
0%
A sparrow in the nest on a tree
0%
A squirrel going up a tree

How much height must an object of mass

$5kg$ be lifted to have potential energy

$5000J$ ?

$[g=10 m/s^2]$

0%
$100m$
0%
$50m$
0%
$120m$
0%
$110m$

Explanation

Gravitation potential energy is given by

$P.E = mgh$

Given:

Mass

$m=5 \ kg$

$g=10 \ m/s^2$

$5000 = 5 \times 10 \times h$

$\implies h = 100 m$

A car is accelerated on a levelled road and attains a speed of

$5$ times its initial speed. In this process , the kinetic energy of the car:

0%
does not change
0%
becomes $4$ times that of initial kinetic energy
0%
becomes $8$ times that of initial kinetic energy
0%
becomes $25$ times that of initial kinetic energy

Explanation

Lets initial velocity of car is $v$ , final velocity of this car will be $5\ v$

$K.E = \dfrac{1}{2}$ $mv ^{2}$

$\therefore \ K.E\propto v^2$

$K.E_{f}=K.E_i \left(\dfrac{v_f}{v_i}\right)^2$ where

$f$ stand for final and

$i$ for initial.

$K.E_{f}=K.E_i \left(\dfrac{5v}{v}\right)^2=25\ K.E_i$

So, when the speed increases to 5 times of its initial velocity, the kinetic energy will become 25 times.

The correct answer is d) becomes

$25$ times that of initial kinetic energy

No work is done when:

0%
a nail is hammered into a wooden box
0%
a box is pushed along horizontal floor
0%
there is no component of force, parallel to the direction of motion
0%
none of these

Explanation

The formula for work done when the force is applied at an angle to the direction of motion is given by,

$W = FS \,cos \theta$
When there will be no component of force, parallel to the direction of motion then force F in direction of motion is zero, so work done will also be zero.

Potential energy of a person is minimum when:

0%
Person is standing
0%
Person is sitting on a chair
0%
Person is sitting on the ground
0%
Person is lying on the ground

Explanation

Potential energy of a body is defined as energy of a body due to its position in gravitational field.

In general,
Potential energy

$= M \times g \times h$ where: h is the height above ground.
If the person will be lying on ground then it will have minimum height above the ground therefore potential energy of the person will also be minimum.

A stone is thrown upwards as shown in the diagram. When it reaches

$P,$ which of the following has the greatest value for the stone?

0%
its acceleration
0%
its kinetic energy
0%
its potential energy
0%
its weight

Write true or false for the following statements:
If we know the speed and mass of an object, we can find out its kinetic energy.

0%
True
0%
False

Explanation

Kinetic energy KE of a body of mass m moving with velocity or speed v is given by the formula:

$KE = \dfrac{1}{2} mv^2$
The formula suggests that kinetic energy depends on mass and speed of the object. So, if we know the speed v and mass m of an object, we can find out its kinetic energy.

The SI unit of work is Watt.

0%
True
0%
False

Explanation

Work done on an object is defined as the magnitude of the force multiplied by the distance moved by the object in the direction of the applied force. The SI unit of work is

$Joule$ .

$1\ joule = 1\ Newton \times 1\ metre$

CBSE Questions for Class 9 Physics Work And Energy Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

CBSE Questions for Class 9 Physics Work And Energy Quiz 12 - MCQExams.com

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Practice Class 9 Physics Quiz Questions and Answers

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