JEE Main Physics Work, Energy And Power Online Test
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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 threetenth of his weight. The work done by the slide on the boy as he comes down is
Correct
Force, F = 3/10 mg
As, W = – Fs
Or W = – 3/10 mgs
Or W = – 3/10 × 200 × 10 J = 600 J
Incorrect
Force, 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
υt graph of an object of mass 1 kg is shown. select the wrong statement.
Correct
Initial velocity = final velocity = 0.
But displacement ≠ 0
From work energy theorem W = ∆KE = 0
Incorrect
Initial velocity = final velocity = 0.
But displacement ≠ 0
From work energy theorem W = ∆KE = 0

Question 3 of 20
3. Question
Velocitytime 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
Correct
From 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
Correct
Area 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
Incorrect
Area 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
Correct
E_{i} – E_{f} = Work done against friction
∴ 1/2 mv^{2} – 1/2 kx^{2} = (μmg)x
Substituting the value we get, x = 1 m
Incorrect
E_{i} – E_{f} = Work done against friction
∴ 1/2 mv^{2} – 1/2 kx^{2} = (μmg)x
Substituting 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?
Correct
As 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.
Incorrect
As 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
Correct
F×du/dx = = 0 at B and C
Incorrect
F×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.
Correct
At 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.
Incorrect
At 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.
Correct
Work 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}) = 250
Area 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
Incorrect
Work 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}) = 250
Area 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:
Correct
Incorrect

Question 12 of 20
12. Question
Velocitytime 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
Correct
Initial velocity of particle, v_{i} = 20 ms^{1}
Final velocity of the particle, v_{f} = 0
According to work energy theorem
W_{net} = ∆KE = K_{f} – K_{i}
= 1/2 × 2(0^{2} – 20^{2}) = – 400 J
Incorrect
Initial velocity of particle, v_{i} = 20 ms^{1}
Final velocity of the particle, v_{f} = 0
According 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})
Correct
x = 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
Incorrect
x = 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 body
S_{3}: Work done by a force between two points is same along many number of paths then it must be a conservative force.
Correct
S_{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
S_{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 is
Correct
∵ 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
Correct
Power 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
Incorrect
Power 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 body
S_{3}: Work done by a force between two points is same along many number of paths then it must be a conservative force.
Correct
S_{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
S_{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 18 of 20
18. Question
The potential energy function for a particle executing linear SHM is given by V(x) = 1/2kx^{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_{m}. If V and K indicate the PE and KE respectively of the particle at x = + x_{m} then which of the following is correct?
Correct
At 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.
Incorrect
At 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
Correct
Incorrect

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
Correct
Initial KE of the body = 1/2 mv^{2} = 1/2 × 25 × 4 = 50 J
Work done against resistive force = Area between Fx graph = 1/2 × 4 × 20 = 40 J
Final KE = Initial KE – Work done against resistive force = 50 – 40 = 10 J
Incorrect
Initial KE of the body = 1/2 mv^{2} = 1/2 × 25 × 4 = 50 J
Work done against resistive force = Area between Fx graph = 1/2 × 4 × 20 = 40 J
Final KE = Initial KE – Work done against resistive force = 50 – 40 = 10 J