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JEE • PHYSICS • WORK-ENERGY-POWER

Work-energy-power practice · Medium

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These sets use single-correct-answer MCQs. They do not cover every JEE Main or JEE Advanced question format and are not full-length mock papers.

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331 Medium
331 of 496
Application +4 | -1
When a belt moves horizontally at a constant speed of 1.5ms−11.5\mathrm{ }m{s^{ - 1}} , gravel is falling on it at 5kgs−15\mathrm{ }kg{s^{ - 1}}. Then the extra power needed to drive the belt is
View answer and explanation
Correct option: D
11.25 W
Option D (11.25 W) is correct because it is the power calculated using the formula P=(dm/dt)V2P = (dm/dt)V^2 where dm/dt is the rate of mass falling and V is the belt speed.
332 Medium
332 of 496
Application +4 | -1
A force of 1200 N acting on a stone by means of rope slides the stone through a distance of 10 m in a direction inclined at 600{60^0} to the force. The work done by the force is [Joules].
View answer and explanation
Correct option: B
6000
Option B (6000) is correct because the work done is calculated using W = F * S * cos(θ).
333 Medium
333 of 496
Application +4 | -1
The power of a pump, which can pump 200 kg of water to a height of 200 m in 10 sec is ( g=10m/s210m/{s^2})
View answer and explanation
Correct option: A
40 kW
Option A (40 kW) is correct because the power is calculated by dividing the potential energy gained by the water by the time taken.
334 Medium
334 of 496
Understanding +4 | -1
In an inelastic collision which of the following does not remain conserved?
View answer and explanation
Correct option: B
kinetic energy
Option B (kinetic energy) is correct because in an inelastic collision, kinetic energy is not conserved, but is converted into other forms of energy like heat or sound.
335 Medium
335 of 496
Understanding +4 | -1
Mud thrown on a wall and sticking to it is an example for:
View answer and explanation
Correct option: D
Perfectly inelastic collision
Option D is correct because in a perfectly inelastic collision, the objects stick together after the collision.
336 Medium
336 of 496
Application +4 | -1
A force moves a block of mass 4 kg4 \mathrm{~kg} on horizontal smooth surface such that the position xx of the block as a function of time tt is given as x=t3x=t^3. Here xx is in metre and tt is in second. Work done by the force in first three second is
View answer and explanation
Correct option: D
1458 J1458 \mathrm{~J}
Option D (1458 J1458 \mathrm{~J}) is correct because the work done is equal to the change in kinetic energy, and the velocity is the derivative of the position function.
337 Medium
337 of 496
Application +4 | -1
A particle of mass mm moves on a straight line with its velocity increasing with distance according to the equation v=αx\mathrm{v}=\alpha \sqrt{\mathrm{x}}, where α\alpha is a constant. The total work done by all the forces applied on the particle during its displacement from x=0\mathrm{x}=0 to x=d\mathrm{x}=\mathrm{d}, will be:
View answer and explanation
Correct option: C
mα2d2\frac{\mathrm{m} \alpha^2 \mathrm{d}}{2}
Option C (mα2d2\frac{{\mathrm{{m}} \alpha^2 \mathrm{{d}}}}{{2}}) is correct because the work done is equal to the change in kinetic energy, which depends on the velocity and displacement.
338 Medium
338 of 496
Application +4 | -1
A body of mass 2 kg2 \mathrm{~kg} moving under a force has relation between displacement xx and time tt as x=t33x=\frac{t^3}{3} where xx is in metre and tt is in sec. The work done by the body in first two second will be
View answer and explanation
Correct option: B
16 joule
Option B (16 joule) is correct because the work done is equal to the change in kinetic energy, which is calculated using the velocity at t=2s.
339 Medium
339 of 496
Application +4 | -1
A body dropped from a height HH reaches the ground with a speed of 1.2gH1.2 \sqrt{g H}. Calculate the work done by air-friction.
View answer and explanation
Correct option: D
−0.28 mgH-0.28 \mathrm{~mgH}
Option D (−0.28 mgH-0.28 \mathrm{~mgH}) is correct because the work done by air friction is the difference between the total work done and the work done by gravity.
340 Medium
340 of 496
Analyzing +4 | -1
An object of mass mm slides down a hill of arbitrary shape and after travelling a certain horizontal path stops because of friction. The total vertical height descended is hh. The friction coefficient is different for different segments for the entire path but is independent of the velocity and direction of motion. The work that a tangential force must perform to return the object to its initial position along the same path is
View answer and explanation
Correct option: D
2mgh2 m g h
Option D (2mgh2 m g h) is correct because the work done against friction during the descent is mghmgh, and the work required to return the object to its initial position is also mghmgh, totaling 2mgh2mgh.
341 Medium
341 of 496
Analyzing +4 | -1
Which of the following statements is correct?
View answer and explanation
Correct option: A
Kinetic energy of a system can be changed without changing its momentum.
Option A (Kinetic energy of a system can be changed without changing its momentum) is correct because internal forces can change kinetic energy while conserving momentum.
342 Medium
342 of 496
Understanding +4 | -1
Roads are banked on curves so that
View answer and explanation
Correct option: A
The speeding vehicles may not fall outwards
Option A is correct because banking roads helps provide the necessary centripetal force to prevent speeding vehicles from skidding outwards.
343 Medium
343 of 496
Application +4 | -1
A bullet fired from a gun with a velocity of 104  ms−1{10^4}\mathrm{\;m}{\mathrm{s}^{ - 1}} goes through a bag full of straw. If the bullet loses half of its kinetic energy in the bag, its velocity when it comes out of the bag will be
View answer and explanation
Correct option: A
7071.06  ms−17071.06\mathrm{\;m}{\mathrm{s}^{ - 1}}
Option A (7071.06 m/s⁻¹) is correct because the velocity is calculated by considering the loss of kinetic energy, resulting in a reduced velocity.
344 Medium
344 of 496
Understanding +4 | -1
Two particles are seen to collide and move jointly together after the collision. During such a collision, for the total system,
View answer and explanation
Correct option: B
linear momentum is conserved but not the mechanical energy.
Option B is correct because in an inelastic collision, linear momentum is always conserved, but kinetic energy is not, as some energy is converted into other forms like heat or sound.
345 Medium
345 of 496
Understanding +4 | -1
When work done on a particle is positive
a) K.E remains constant
b) Momentum increases
c) K.E increases
d) K.E decreases
View answer and explanation
Correct option: B
b, c
Option B is correct because positive work done on a particle increases its kinetic energy, and consequently, its momentum also increases.
346 Medium
346 of 496
Analyzing +4 | -1
The slope of kinetic energy and displacement curve for a particle in motion will be
View answer and explanation
Correct option: B
directly proportional to the acceleration of the particle
Option B is correct because the slope of the kinetic energy and displacement curve is directly proportional to the acceleration of the particle.
347 Medium
347 of 496
Application +4 | -1
A particle moves along x-axis from x=1m to x=3m under the effect of the force F=3x2−2x+5NF = 3{x^2} - 2x + 5\mathrm{ }N work done in process is
View answer and explanation
Correct option: D
28 J
Option D (28 J) is correct because it is the result of integrating the force function over the given displacement range.
348 Medium
348 of 496
Application +4 | -1
Two spherical bodies of the same mass M are moving with velocities v1 and v2{v_1}\,and\,{v_2} .These collide perfectly inelastically , then the loss in kinetic energy is
View answer and explanation
Correct option: C
14M(v1−v2)2\frac{1}{4}M{\left( {{v_1} - {v_2}} \right)^2}
Option C (14M(v1−v2)2\frac{1}{4}M{\left( {{v_1} - {v_2}} \right)^2}) is correct because it represents the kinetic energy lost during a perfectly inelastic collision of two equal masses.
349 Medium
349 of 496
Application +4 | -1
The potential energy of 1kg1kg particle free to move along the x−axisx - axis is given by U (x)=[x44−x22]JU\,(x) = \left[ {\frac{{{x^4}}}{4} - \frac{{{x^2}}}{2}} \right]J. The total mechanical energy of the particle is 2J2J. What is the maximum speed (in  ms−1)\left( {in\,\,m{s^{ - 1}}} \right) of the particle?
View answer and explanation
Correct option: A
32ms−1\frac{3}{{\sqrt 2 }}m{s^{ - 1}}
Option A (32ms−1\frac{3}{{\sqrt 2 }}m{s^{ - 1}}) is correct because it is calculated by finding the minimum potential energy and using the conservation of energy to determine the maximum kinetic energy and hence the maximum speed.
350 Medium
350 of 496
Application +4 | -1
A body of mass m{m} attached to a thread is revolved along a vertical circle of radius rr. If its velocity at the topmost point of the circle is vv, the tension of the thread at the instant will be :
View answer and explanation
Correct option: C
(mv/r)−mg({mv} / {r})-{mg}
Option C ((mv/r)−mgmv / {r})-{mg}) is correct because it represents the net force required to provide the centripetal acceleration at the topmost point.
351 Medium
351 of 496
Application +4 | -1
The position x of a particle of mass 2 kg moving along straight line is given as x=(t−2)2x = {\left( {t - 2} \right)^2}, where t is in second and x is in metre. The work done by the force in first 2s is
View answer and explanation
Correct option: B
-16 J
Option B (-16 J) is correct because the work done is equal to the change in kinetic energy, which is negative in this case.
352 Medium
352 of 496
Analyzing +4 | -1
Assertion (A): Collision between two particles is not necessarly associated with physical contact between them
Reason(R): Only in physical contact, momentum transfer takes place
View answer and explanation
Correct option: C
If Assertion is true but Reason is false
Option C (If Assertion is true but Reason is false) is correct because collisions don't require physical contact, but momentum transfer still occurs.
353 Medium
353 of 496
Understanding +4 | -1
If a particle of mass m1m_1 approaches another particle of mass m2m_2, initially at rest, with speed vvv_v, the speed of the cm\mathrm{cm} of the system is vcm=m1v1m1+m2v_{\mathrm{cm}}=\frac{m_1 v_1}{m_1+m_2} before collision takes place. If the collision is elastic, the speed of the cm\mathrm{cm} of the system after the collision is
View answer and explanation
Correct option: B
m1v1m1+m2\frac{m_1 v_1}{m_1+m_2}
Option B (m1v1m1+m2\frac{{m_1 v_1}}{{m_1+m_2}}) is correct because in an elastic collision with no external forces, the velocity of the center of mass remains constant.
354 Medium
354 of 496
Understanding +4 | -1
Which of the following forces does not act on a body moving in uniform circular motion?
View answer and explanation
Correct option: A
Centripetal force
Option A (Centripetal force) is correct because centripetal force is not a fundamental force but rather the net force causing circular motion, provided by other forces.
355 Medium
355 of 496
Application +4 | -1
A ball is spun with angular acceleration α=6t2−2t\alpha = 6{t^2} - 2t, where tt is in second and α\alpha is in rads−2rad{s^{ - 2}}. At t=0t = 0, the ball has angular velocity of 10 rads−110\,rad{s^{ - 1}} and angular position of 4 rad. The most appropriate expression for the angular position of the ball is:
View answer and explanation
Correct option: B
t42−t33+10t+4\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4
Option B (t42−t33+10t+4\frac{{{t^4}}}{2} - \frac{{{t^3}}}{3} + 10t + 4) is correct because it is obtained by integrating the angular acceleration twice with respect to time, using the given initial conditions for angular velocity and angular position.
356 Medium
356 of 496
Application +4 | -1
A solid sphere of mass 2 kg2 \mathrm{~kg} is rolling on a frictionless horizontal surface with velocity 6 m/s\mathrm{m} / \mathrm{s}. It collides on the free end of an ideal spring whose other end is fixed. The maximum compression produced in the spring will be (Force constant of the spring =36 N/m=36 \mathrm{~N} / \mathrm{m}).
View answer and explanation
Correct option: B
2.8 m\sqrt{2.8} \mathrm{~m}
Option B (2.8 m\sqrt{2.8} \mathrm{~m}) is correct because it's the maximum compression calculated by equating the kinetic energy of the rolling sphere to the potential energy stored in the spring.
357 Medium
357 of 496
Analyzing +4 | -1
A body is acted upon by a constant force directed towards a fixed point. The magnitude of the force varies inversely as the square of the distance from the fixed point. What is the nature of the path?
View answer and explanation
Correct option: C
Circle
Option C is correct because a force directed towards a fixed point, with magnitude inversely proportional to the square of the distance, results in circular motion.
358 Medium
358 of 496
Understanding +4 | -1
Assertion(A): If work done by conservative force is negative, then potential energy associated with that force should increase.

Reason(R): This is according to the relation, ΔU=−W\Delta U=-W. Here, ΔU\Delta U is change in potential energy and WW is work done by conservative force.
View answer and explanation
Correct option: A
If both Assertion and Reason are correct and Reason is the correct explanation of Assertion.
The correct answer is A (If both Assertion and Reason are correct and Reason is the correct explanation of Assertion) because the change in potential energy is the negative of the work done by a conservative force.
359 Medium
359 of 496
Understanding +4 | -1
A ball hits a floor and rebounds after an inelastic collision. In this case,
a) Total energy of the ball and the earth remains the same
b) Total momentum of the ball and the earth is conserved
c) Momentum of the ball just after the collision is same as that just before the collision
d) Mechanical energy of the ball remains the same during the collision
View answer and explanation
Correct option: C
a, b
Option C is correct because in an inelastic collision, the total energy of the ball and earth remains the same, and the total momentum of the ball and earth is conserved.
360 Medium
360 of 496
Application +4 | -1
A projectile is fired from the top of a 40 m40 \mathrm{~m} high cliff with an initial speed of 50 m/s50 \mathrm{~m} / \mathrm{s} at an unknown angle. Find its speed when it hits the ground.
View answer and explanation
Correct option: A
58 m/s58 \mathrm{~m} / \mathrm{s}
Option A (58 m/s58 \mathrm{~m} / \mathrm{s}) is correct because the final speed is calculated using the conservation of energy principle.
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