# JEE Main Physics Work, Energy And Power Online Test, JEE Main Mock Test

## JEE Main Physics Work, Energy And Power Online Test

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JEE Main Physics Work, Energy And Power Online Test. JEE Main Online Test for Physics Work, Energy And Power. JEE Main Full Online Quiz **for Physics Work, Energy And Power**. **JEE Main Free Mock Test Paper 2018.** JEE Main 2018 Free Online Practice Test, Take JEE Online Test for All Subjects. JEE Main Question and Answers for Physics Work, Energy And Power. In this test You may find JEE Main all subjects Questions with Answers. Check JEE Main Question and Answers in English. This mock Test is free for All Students. Mock Test Papers are very helpful for Exam Purpose, by using below Mock Test Paper you may Test your Study for Next Upcoming Exams. Now Scroll down below n **Take JEE Main Physics Work, Energy And Power…**

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- Question 1 of 20
##### 1. Question

In a children’s park, there is a slide which has a total length of 10 m and a height of 8.0 m. A vertical ladder is provided to reach the top. A boy weighing 200 N climbs up the ladder to the top of the slide and slides down to the ground. the average friction offered by the slide is three-tenth of his weight. The work done by the slide on the boy as he comes down is

CorrectForce, F = 3/10 mg

As, W = – Fs

Or W = – 3/10 mgs

Or W = – 3/10 × 200 × 10 J = 600 J

IncorrectForce, F = 3/10 mg

As, W = – Fs

Or W = – 3/10 mgs

Or W = – 3/10 × 200 × 10 J = 600 J

- Question 2 of 20
##### 2. Question

υ-t graph of an object of mass 1 kg is shown. select the

**wrong**statement.CorrectInitial velocity = final velocity = 0.

But displacement ≠ 0

From work energy theorem W = ∆KE = 0

IncorrectInitial velocity = final velocity = 0.

But displacement ≠ 0

From work energy theorem W = ∆KE = 0

- Question 3 of 20
##### 3. Question

Velocity-time graph of a particle moving in a straight line is as shown in figure. Mass of the particle is 2 kg. Work done by all the forces acting on the particle in time interval between t = 0 to t = 10 s is

CorrectFrom work, energy theorem, W = ∆KE

= K

_{f}– K_{i}= 1/2 m= 1/2 (2) (400 – 100) = 300 J

Incorrect - Question 4 of 20
##### 4. Question

The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. the mass of the body is 25 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be

CorrectArea under curve = 1/2 (4) (20) = 40 J

W = work done by resistive force F = – 40 J

– 40 = K

_{f}– K_{i}, K_{i}= 50 J, so, K_{f}= 50 – 40 = 10 JIncorrectArea under curve = 1/2 (4) (20) = 40 J

W = work done by resistive force F = – 40 J

– 40 = K

_{f}– K_{i}, K_{i}= 50 J, so, K_{f}= 50 – 40 = 10 J - Question 5 of 20
##### 5. Question

A block of mass 50 kg is projected horizontally on a rough horizontal floor. The coefficient of friction between the block and the floor is 0.1. The block strikes a light spring of stiffness k = 100 N/m with a velocity 2 m/s, the maximum compression of the spring is

CorrectE

_{i}– E_{f}= Work done against friction∴ 1/2 mv

^{2}– 1/2 kx^{2}= (μmg)xSubstituting the value we get, x = 1 m

IncorrectE

_{i}– E_{f}= Work done against friction∴ 1/2 mv

^{2}– 1/2 kx^{2}= (μmg)xSubstituting the value we get, x = 1 m

- Question 6 of 20
##### 6. Question

Two inclined frictionless tracks, one gradual and the other steep meet at A from where two stones are allowed to slide down from rest, one on each track as shown in figure.

Which of the following statement is correct?

CorrectAs both surface I and II are frictionless and two stones slide from the same height, therefore, both the stone reach the bottom with same speed 1/2. As acceleration down plane II is larger (a

_{2}= g sin θ_{2}greater than a_{1}= g sin θ_{1}), therefore, stone II reaches the bottom earlier than stone I.IncorrectAs both surface I and II are frictionless and two stones slide from the same height, therefore, both the stone reach the bottom with same speed 1/2. As acceleration down plane II is larger (a

_{2}= g sin θ_{2}greater than a_{1}= g sin θ_{1}), therefore, stone II reaches the bottom earlier than stone I. - Question 7 of 20
##### 7. Question

The potential energy of a particle varies with distance x as shown in the graph..

The force acting on the particle is zero at

CorrectF×du/dx = = 0 at B and C

IncorrectF×du/dx = = 0 at B and C

- Question 8 of 20
##### 8. Question

In the given curved road, if particle is released from A, then

Correct- If the surface is smooth, then the kinetic energy at B never be zero.
- If the surface is rough, the kinetic energy at B be zero. Because, work done by force of friction is negative. If work done by friction is equal to mgh then, net work done on ody will be zero. Hence, net change in kinetic energy is zero. Hence, (A) is correct.
- If the surface is rough, the kinetic energy at B must be lesser than mgh. If surface is smooth, the kinetic energy at B is equal to mgh.
- The reason is same as in (B) and (A)

Incorrect- If the surface is smooth, then the kinetic energy at B never be zero.
- If the surface is rough, the kinetic energy at B be zero. Because, work done by force of friction is negative. If work done by friction is equal to mgh then, net work done on ody will be zero. Hence, net change in kinetic energy is zero. Hence, (A) is correct.
- If the surface is rough, the kinetic energy at B must be lesser than mgh. If surface is smooth, the kinetic energy at B is equal to mgh.
- The reason is same as in (B) and (A)

- Question 9 of 20
##### 9. Question

The given plot shows the variation of U, the potential energy of interaction between two particles with the distance separating them, r. then which of the following statements are correct.

CorrectAt point ‘C’, the potential energy is minimum, hence it is a point of stable equilibrium. Also, from E to F, the slope is negative i.e., DU/dr < 0

Hence, the force of interaction between the particles is repulsive between points E and F.

IncorrectAt point ‘C’, the potential energy is minimum, hence it is a point of stable equilibrium. Also, from E to F, the slope is negative i.e., DU/dr < 0

Hence, the force of interaction between the particles is repulsive between points E and F.

- Question 10 of 20
##### 10. Question

A body of mass 5 kg is acted upon by a variable force. The force varies with the distance covered by the body. Find the kinetic energy of the body when the body has covered 30 m distance? Assume that the body starts from rest.

CorrectWork done by force when body has covered 25 m

= Area under curve upto distance 25 m

Now W = ∆K = 1/2 × 5 (v

^{2}– 0^{2}) = 250Area between 25 m to 30 m (1/2)5 × 18 = 45

So, total work done by variable force, till 30 m is 250 + 45 = 295 Joule

So change in kinetic energy = 295 J

So, final kinetic energy = 295 J

IncorrectWork done by force when body has covered 25 m

= Area under curve upto distance 25 m

Now W = ∆K = 1/2 × 5 (v

^{2}– 0^{2}) = 250Area between 25 m to 30 m (1/2)5 × 18 = 45

So, total work done by variable force, till 30 m is 250 + 45 = 295 Joule

So change in kinetic energy = 295 J

So, final kinetic energy = 295 J

- Question 11 of 20
##### 11. Question

The potential energy of a system increases if work is done:

CorrectIncorrect - Question 12 of 20
##### 12. Question

Velocity-time graph of a particle of mass 2 kg moving in a straight line is as shown in figure. Work done by all force on the particle is

CorrectInitial velocity of particle, v

_{i}= 20 ms^{-1}Final velocity of the particle, v

_{f}= 0According to work energy theorem

W

_{net}= ∆KE = K_{f}– K_{i}= 1/2 × 2(0

^{2}– 20^{2}) = – 400 JIncorrectInitial velocity of particle, v

_{i}= 20 ms^{-1}Final velocity of the particle, v

_{f}= 0According to work energy theorem

W

_{net}= ∆KE = K_{f}– K_{i}= 1/2 × 2(0

^{2}– 20^{2}) = – 400 J - Question 13 of 20
##### 13. Question

A vertical spring of force constant 100 N/m is attached with a hanging mass of 10 kg. Now an external force is applied on the mass so that the spring is stretched by additional 2m. The work done by the force F is (g = 10 m/s

^{2})Correctx = elongation in spring due to mass 10 kg =

10×10/100 = 1 m

W

_{F}= 1/2 × 100 × [(3)^{2}– (1)^{2}] – 10 × 10 × 2 = 200 JIncorrectx = elongation in spring due to mass 10 kg =

10×10/100 = 1 m

W

_{F}= 1/2 × 100 × [(3)^{2}– (1)^{2}] – 10 × 10 × 2 = 200 J - Question 14 of 20
##### 14. Question

S

_{1}: For a body moving under action of certain forces, one of the force may do negative work event if the kinetic energy of the body is increasing.S

_{2}: A net force that does not work on the body can change velocity of that bodyS

_{3}: Work done by a force between two points is same along many number of paths then it must be a conservative force.CorrectS

_{1}: The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing.S

_{2}: A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity.S

_{3}: For a force to be conservative work done by force should be equal on each path.IncorrectS

_{1}: The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing.S

_{2}: A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity.S

_{3}: For a force to be conservative work done by force should be equal on each path. - Question 15 of 20
##### 15. Question

A body of mass M is moving with a uniform speed of 10 m/s on frictionless surface under the influence of two force F

_{1}and F_{2}. The net power of the system isCorrect∵ Speed is constant.

∴ Work done by force = 0

∴ Power = 0

Incorrect∵ Speed is constant.

∴ Work done by force = 0

∴ Power = 0

- Question 16 of 20
##### 16. Question

A man is supplying a constant power of 500 J/s to a massless string by pulling it at a constant speed of 10 m/s as shown. it is known that kinetic energy of the block is increasing at a rate of 100 J/s. Then, the mass of the block is

CorrectPower supplied = increase in kinetic energy + potential energy per second.

Speed is 10 m/s Therefore, in 1 s mass m rises a height of 10 m.

∴ 500 = 100 + (m × 10 × 10) = m = 4 kg

IncorrectPower supplied = increase in kinetic energy + potential energy per second.

Speed is 10 m/s Therefore, in 1 s mass m rises a height of 10 m.

∴ 500 = 100 + (m × 10 × 10) = m = 4 kg

- Question 17 of 20
##### 17. Question

S

_{1}: For a body moving under action of certain forces, one of the force may do negative work event if the kinetic energy of the body is increasing.S

_{2}: A net force that does not work on the body can change velocity of that bodyS

_{3}: Work done by a force between two points is same along many number of paths then it must be a conservative force.CorrectS

_{1}: The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing.S

_{2}: A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity.S

_{3}: For a force to be conservative work done by force should be equal on each path.Incorrect_{1}: The net force on the body may have acute angle with its velocity, but one of the constituent force may have obtuse angle with the velocity. Such a force shall perform negative work on the body even though the kinetic energy of the body is increasing._{2}: A net force that is always perpendicular to velocity of the particle does no work but changes the direction of its velocity.S

_{3}: For a force to be conservative work done by force should be equal on each path. - Question 18 of 20
##### 18. Question

The potential energy function for a particle executing linear SHM is given by V(

*x*) = 1/2*kx*^{2}where*k*is the force constant of the oscillator. For*k*= 0.5 N/m, the graph of V(*x*)*uersus x*is shown in the figure. A particle of total energy E turns back when it reaches*x*=__+__*x*. If V and K indicate the PE and KE respectively of the particle at_{m}*x*=__+__*x*_{m}then which of the following is correct?CorrectAt x = + x

_{m}, the particle turns back. Therefore, its velocity at this point is zero. Therefore, kinetic energy K = 0. The total energy E is in the form of potential energy i.e., V = E.IncorrectAt x = + x

_{m}, the particle turns back. Therefore, its velocity at this point is zero. Therefore, kinetic energy K = 0. The total energy E is in the form of potential energy i.e., V = E. - Question 19 of 20
##### 19. Question

The kinetic energy of a particle continuously increases with time

CorrectIncorrect - Question 20 of 20
##### 20. Question

The graph between the resistive force F acting on a body and the distance covered by the body is shown in the figure. the mass of the body is 2.5 kg and initial velocity is 2 m/s. When the distance covered by the body is 4 m, its kinetic energy would be

CorrectInitial KE of the body = 1/2 mv

^{2}= 1/2 × 25 × 4 = 50 JWork done against resistive force = Area between F-x graph = 1/2 × 4 × 20 = 40 J

Final KE = Initial KE – Work done against resistive force = 50 – 40 = 10 J

IncorrectInitial KE of the body = 1/2 mv

^{2}= 1/2 × 25 × 4 = 50 JWork done against resistive force = Area between F-x graph = 1/2 × 4 × 20 = 40 J

Final KE = Initial KE – Work done against resistive force = 50 – 40 = 10 J