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CBSE Questions for Class 9 Physics Work And Energy Quiz 10 - MCQExams.com
CBSE
Class 9 Physics
Work And Energy
Quiz 10
Calculate the work done by the person to push a box with a force of $$20 N$$ over a distance of $$8.0 m$$ in the direction of the force?
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0%
$$1.6 J$$
0%
$$16 J$$
0%
$$160 J$$
0%
$$1600 J$$
0%
$$16000 J$$
Explanation
Work done $$(W)$$
$$W = F\times S$$
$$F$$; applied force
$$S$$; displacement
Given :
$$F = 20N$$
$$S = 8.0$$ m
$$\Rightarrow W = 20 \times 8.0$$
$$\Rightarrow W = 160\ J$$
Identify the wrong statement about work.
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It is the scalar product of force and displacement
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It is a vector that is always in the same direction as the force.
0%
It is measured in joules.
0%
It has the same units as energy.
Explanation
Work done by a force is given by dot product of force and displacement $$W = F.S =F S$$ $$cos\theta$$
Work done is a scalar quantity.
It is measured in joules and thus has the same units as energy.
Hence statement given in option B is wrong.
How much work does a person do in pushing a box with a force of 10 N over a distance of 8.0 m in the direction of the force?
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0.8 J
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8.0 J
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80 J
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800 J
0%
8000 J
Explanation
The work done$$(W)$$ by a force-
$$W = \text{force} \times \text{displacement}$$
$$F = 10\ N$$
$$\text{dispalcement} = 8\ m$$
$$\Rightarrow W = 10 \times 8$$
$$\Rightarrow W = 80\ J$$
Which of the following is NOT a unit of energy?
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Watt
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eV
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Kilowatt hour
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Joule
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Calorie
Explanation
Hint: Energy and work have same units.
Explanation
:-
$$\bullet$$ Joule, Calorie ,Kilowatt hour and eV are the units of energy while ‘watt’ is the units of power
$$\bullet$$ $$P=\frac{w}{ t} =\frac {joule}{time } =Watt$$
$$\bullet$$ $$ watt \times$$ unit of time is equal to unit of energy ,so Kilowatt hour is unit of energy
$$\textbf{Hence option A correct}$$
Identify the best definition of energy?
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anything that has mass and takes up space
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distance moved in a period of time
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ability to do work
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mode of opperation
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moment of an object with application of force
Explanation
Energy is defined as the ability to do work and its SI unit is Joule.
A body of mass 6 kg is under a force of 6 N which causes displacement in it given by $$S =\frac{t^2}{4}$$ where 't' is time in seconds. The work done by the force in 2 s is:
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0%
12 J
0%
9 J
0%
6 J
0%
3 J
Explanation
$$time = t = 2s$$
$$ displacement = S=\dfrac{t^2}{4}=\dfrac{2^2}{4}=1\ m $$
The work done,
$$W= force\times displacement =F \times S=6 \times 1=6\ J$$
Person $$A$$ stands on the right side of a cart, while Person $$B$$ stands on the left side. The cart is initially at rest. Person $$A$$ pushes on the cart with a force of $$22.0 N$$, and Person $$B$$ pushes on the cart with a force of $$33.0 N$$.
The cart, with no other forces acting on it, starts to move and rolls through a distance of $$2.0 m$$ while both people continue to push with their respective forces.
How much work was done on the cart by Person $$B$$?
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$$-44.0 J$$
0%
$$+22.0 J$$
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$$+44.0 J$$
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$$+66.0 J$$
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None of the above
Explanation
Force exerted by person B $$F_B = 33.0$$ N
Displacement of the box in the direction of $$F_B$$ $$S = 2.0$$ m
$$\therefore$$ Work done by person B $$W_B = F_B\times S = 33.0 \times 2.0 = + 66.0$$ J
Define one watt:
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one joule per second
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one second per joule
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ten joules per second
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voltage divided by current
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resistance multiplied by voltage
Explanation
Watt is a SI unit of power which is defined as the energy per unit time i.e $$P =\dfrac{E}{t}$$
Hence one watt is defined as one joule of energy used in one second.
$$\therefore$$ $$W = \dfrac{J}{s}$$
A block of mass 10 kg is pulled by a constant horizontal force of 19 N and it is displaced by 15 m across the floor. Calculate the work done.
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0%
1.3 J
0%
30 J
0%
285 J
0%
5586 J
0%
1J
Explanation
Given : Force $$F =19 N$$
Displacement $$S= 15$$ m
Work done by a force is $$W= F S $$
$$\therefore$$ $$W = 19 \times 15 = 285$$ $$J$$
How much work does a person do in pushing a box with a force of $$20\ N$$ over a distance of $$8.0\ m$$ in the direction of the force?
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0%
$$1.6\ J$$
0%
$$16\ J$$
0%
$$160\ J$$
0%
$$1600\ J$$
0%
$$16000\ J$$
Explanation
Given,
Force applied, $$F = 20\ N$$
Displacement, $$d=+8.0\ m$$ (in the direction of the force)
Work done, $$W$$
$$W=F\times d$$
$$W=20\times 8$$
$$W=160\ J$$
Does the Law of Conservation of energy apply to machines?
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No, as machines require a constant input of energy in order to work
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Yes, the energy required to maintain motion is balanced by the energy lost to friction
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No, as machines can create energy from motion
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Yes, but only to perpetual motion machines
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No, as the law would apply to only perpetual motion machines
Explanation
Machines require input work to be done on them, which increases the potential, or kinetic energy, or both, of the system.
However this work is done on the machines by external systems that utilize their energies to convert them to work.
So even though Law of Conservation of Energy holds for the entire system as a whole, it does not stand for the machines alone.
Identify which of the following option best describe work.
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a form of energy
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measured in Coulombs
0%
the potential charge of a particle
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equal to the voltage
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the same for all charge collections
Explanation
The energy possessed by an object is measured in terms of its capacity of doing work.Work done is converted into energy.This work done is given by:
$$WORK=FORCE\times DISPLACEMENT$$
Hence, work is a form of energy.It's SI unit is same as energy , that is $$Joule$$.
A man with a box on his head is climbing up a ladder. What work is said to be done by the man on the box?
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0%
Positive
0%
Negative
0%
Zero
0%
Undefined
Explanation
When man is climbing up a ladder by keeping box in hand, its mean that Box is also moving up. During upward movement of Box, man will apply constant force on the Box in upward direction which is just greater then the weight of Box.
Because Direction of force on the box and direction of displacement of box is in same direction, So Work done by man on the box will be positive.
A $$3.0\ N$$ force applied on a $$0.40\ kg$$ mass makes the mass move at a constant speed of $$3.0\ {m}{s^{-1}}$$, held back by friction. In the span of it moving through $$4.0\ m$$, the net work done on the mass is:
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0%
$$0\ J$$
0%
$$0.90\ J$$
0%
$$12\ J$$
0%
$$-12\ J$$
0%
$$-0.90\ J$$
Explanation
As the mass is moving with a constant speed, thus net external force acting on the mass is zero.
$$F_{net} = 0$$
Displacement of the mass
$$S = 4.0$$ m
Net work done on the mass
$$W_{net} = F_{net} \times S$$
$$\Rightarrow W_{net} = 0\ J$$
When a stone tied to a string is whirled in a circle, what kind of work is said to be done on it by the string?
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0%
Positive
0%
Negative
0%
Zero
0%
Undefined
Explanation
When a stone is tied to a string and whirled in circle $$no\ work$$ is said to be done.
The string and tangent of the circle will be $$perpendicular.$$
Hence, work done is zero.
SI unit of energy is _________ .
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0%
newton
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joule
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erg
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metre
Explanation
Joule is the SI unit of energy.
The running water primarily possesses potential energy.
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0%
True
0%
False
Explanation
The energy of an object by virtue of its motion is known as kinetic energy. Running water possesses kinetic energy.
The energy possessed by a body due to its change in position
or shape is called potential energy.
Find the work done by a force of $$5 N$$ in displacing a book through $$20 cm$$ along the direction of the force.
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0%
$$1 J$$
0%
$$3 J$$
0%
$$4 J$$
0%
$$5 J$$
Explanation
Given:
Force $$F=5\ N$$
Displacement $$ s = 20\ cm = 0.2 \ m $$
Work done = $$Force(F)\times displacement(s) $$
= $$5 \times 0.2 $$
= $$1\ J$$
The work done is positive as the book moves in the direction of the force.
How much work is done in ergs in pulling a box $$2\ m$$ across a tabletop with a force of $$20\ N$$ in the horizontal direction?
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0%
$$40$$
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$$4000$$
0%
$$4\times 10^7$$
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$$40\times 10^7$$
Explanation
Given,
Displacement, $$s=2\ m$$
Force applied, $$F=20\ N$$
Work done, $$W=F\times s$$
$$W=20\times 2$$
$$W=40\ J$$
We know,
$$1\ J=10^7\ ergs$$
$$\therefore W=40\times 10^7$$
$$W=40\times 10^7\ ergs$$
1 horse power is equal to ______ watt :
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0%
746
0%
500
0%
646
0%
700
Explanation
One horse power (HP) $$=550$$ foot pound per second
Also,
$$1 $$ foot $$=0.3048 \ m$$
$$1$$ pound $$=4.448 \ N $$
So, $$1 \ HP=550\times 0.3048\times 4.448 \ Nms^{-1} \simeq 746 \ Nms^{-1} $$
Since $$ 1 W=1 \ Nms^{-1}$$,
$$\Rightarrow 746 \ Nms^{-1}=746 \ watt$$
Option A is correct.
A boy carrying a box on his head is walking on a level road from one place to another. The boy is doing no work.
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0%
True
0%
False
Explanation
The boy is walking i.e., displacement is in the horizontal direction, and the force acting on the box is due to the gravitational pull in the downward direction. So the displacement and the force are perpendicular to each other.
Hence, the work done is zero.
An air conditioner is rated 240 V, 1.5 kW. The air conditioner is switched on 8 hours each day. What is electrical energy consumed in 30 days?
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0%
360 kWh
0%
2.88 kWh
0%
120 kWh
0%
240 kWh
Explanation
Given,
Power of air conditioner $$P=1.5kW$$
Time to use per day $$t=8\ hr$$
Time to use in 30 days $$T=8\times 30\ hr$$
Energy consumed $$E=Power\times time =1.5\ kW\times 8\times 30\ hr=360kWh $$
Option B
The correct formula to find the velocity of a body with kinetic energy '$$k$$' is:
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$$V=\dfrac{2k}{m}$$
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$$V=\sqrt{\dfrac{2k}{m}}$$
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$$V=\dfrac{4{k}^{2}}{m}$$
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$$V=\dfrac{1}{2}km$$
Explanation
Kinetic energy of a body is given by
$$k=\dfrac{1}{2}mv^2$$
$$v^2$$
= $$\dfrac{2k}{m}$$
$$v=\sqrt {\dfrac{2k}{m}}$$.
Hence, answer is option B.
A $$2kg$$ object is moving at $$3m/s$$. A $$4N$$ force is applied in the direction of motion and then removed
after the object has travelled an additional $$5m$$. The work done by this force is:
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0%
$$10 J$$
0%
$$15 J$$
0%
$$20 J$$
0%
$$25 J$$
Explanation
intial velocity $$u=3m/s$$
final velocity $$v^2=u^2+2as=3^2+(2 \times 2 \times 5)=29$$
$$\implies v=5.3m/s$$
$$KE=\dfrac{1}{2}m(v^2-u^2)=\dfrac{1}{2} \times 2 \times {((5.3)^2-3^2)}=20J$$
$$1$$ kilowatt hr $$=$$ __________ joules.
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0%
$$10.6 \times {10}^{6}$$
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$$3.6 \times {10}^{6}$$
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$$30.6 \times {10}^{6}$$
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$$3.6 \times {10}^{5}$$
Explanation
$$kWh$$ is the commercial unit of energy.
$$1\ kWh =1\times 1000\times 3600 = 3.6\times 10^6$$ joules
$$1 Wh$$ (Watt hour) is equal to :
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0%
$$36\times { 10 }^{ 5 }J$$
0%
$$36\times { 10 }^{ 4 }J$$
0%
$$3600 J$$
0%
$$3600 { Js }^{ -1 }$$
Explanation
Watt is a unit of power such that $$1W =1\dfrac{J}{s}$$
Also we know $$1h = 3600$$ $$s$$
$$\therefore$$ $$1Wh = 1\dfrac{J}{s} \times 3600 $$ $$s = 3600$$ $$J$$
Kinetic energy of a body depends upon its:
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mass
0%
velocity
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density
0%
both A and B
Explanation
Kinetic energy of the body $$K = \dfrac{1}{2}mv^2$$
$$\implies$$ $$K \propto m$$ and $$K\propto v^2$$
Thus kinetic energy of a body depends on the mass of the body as well as its velocity.
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0%
Both Assertion and Reason are correct and Reason is the correct explanation for Assertion
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Both Assertion and Reason are correct but Reason is not the correct explanation for Assertion
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Assertion is incorrect but Reason is correct
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Both Assertion and Reason are incorrect
Explanation
Energy $$E = Pt$$
where $$P$$ is the power and $$t$$ is the time of operation.
Since $$P$$ has unit watt and $$t$$ has a unit hour.
Thus watt-hour is the unit of energy.
Similarly, a kilowatt-hour is the unit of energy.
Thus Assertion is incorrect and Reason is correct.
A bicyclist comes to a skidding stop in $$10 m$$. During this process, the force on the bicycle due to the road is $$200N$$ and is directly opposed to the motion. The work done by the cycle on the road is
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0%
$$+ 2000J$$
0%
$$200J$$
0%
$$zero$$
0%
$$20,000J$$
Explanation
Work done by the road on the cycle is
$$W =F \times s=- 200N \times 10m=-2000J$$
Negative sign indicates that the direction of the force and the displacement is opposite.
The cycle applies an equal and opposite force on the road. But as the displacement of the road is zero,
work done by the cycle on the road is
$$zero$$.
A block of weight $$W$$ is pulled a distance $$l$$ along a horizontal table. The work done by the weight is
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0%
$$Wl$$
0%
$$Wgl$$
0%
$$Wl/g$$
0%
zero
Explanation
Here the weight will act in a vertically downward direction but the displacement is in the horizontal direction. They are perpendicular to each other.
Since, there is no displacement in the direction of the weight W,
work done by the weight (W) will be zero.
Hence the correct answer is option $$D $$.
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