Previous year questions: Work, Energy and Power

Ten questions follow from the official JEE Main 2026 papers and the NCERT Exemplar. Each heading gives the paper and the marks. Attempt each question on paper before reading the answer, and write the energy equation before substituting any number.

Q1 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A 1 kg block is subjected to two simultaneous forces, (2i + 3j + 4k) N and (3i - j - 2k) N, and is moved 25 m along the (3i - 4j) direction. The work done in the process is

Answer: Add the forces to get the net force, 5i + 2j + 2k N. The unit vector along the displacement is (3i - 4j)/5, so the displacement vector is 25 times that, which is 15i - 20j m. The work is the scalar product of force and displacement: 5 times 15 plus 2 times minus 20 plus 2 times zero, which is 75 minus 40, so 35 J.

Hinglish: Forces add karke net force 5i + 2j + 2k. Displacement direction (3i-4j)/5, isliye 25 m ka displacement 15i - 20j. Work = 5x15 + 2x(-20) = 75 - 40 = 35 J.

Q2 (JEE Main 2026, 4 Apr Shift 1, 4 marks)

A body of mass 2 kg begins to move from rest under a time dependent force F = (2t i + 6t squared j) N. The power produced by the force at t = 2 s is

Answer: Force is the rate of change of momentum, so integrate the force from zero to t to get momentum: t squared i + 2 t cubed j. At t = 2 the momentum is 4i + 16j, and dividing by the mass 2 gives velocity 2i + 8j m per second. The force at t = 2 is 4i + 24j N. Power is the scalar product of force and velocity: 4 times 2 plus 24 times 8, which is 8 plus 192, so 200 W.

Hinglish: Force momentum ka rate hai, isliye integrate karke p = t^2 i + 2t^3 j. t = 2 par p = 4i + 16j, velocity p/2 = 2i + 8j. Force at t = 2: 4i + 24j. Power = F·v = 8 + 192 = 200 W.

Q3 (JEE Main 2026, 6 Apr Shift 2, 4 marks)

A body of mass 1 kg moves along a straight line with a velocity v = 2x squared, where x is in metre. The work done by the body during the displacement from x = 0 to x = 5 m is

Answer: Use the work-energy theorem, because the force is not given directly. At x = 0 the velocity is zero and at x = 5 it is 2 times 25, that is 50 m per second. The kinetic energy gained is half of 1 times 2500, which is 1250 J, and that is the work done.

Hinglish: Force direct nahi diya, isliye work-energy theorem use karo. x = 0 par v = 0, aur x = 5 par v = 50 m/s. Kinetic energy 1/2 x 1 x 2500 = 1250 J, matlab work 1250 J.

Q4 (JEE Main 2026, 5 Apr Shift 2, 4 marks)

A mass of 1 kg is kept on an inclined plane of inclination 30 degrees and is at rest relative to it. The whole assembly is then moved vertically upward at a constant velocity of 4 m per second. The work done by the frictional force on the mass in 2 s is (take g = 10 m s^-2)

Answer: Since the mass does not slide on the plane, static friction balances the component of the weight along the slope, so its magnitude is mg sin 30, which is 5 N. The friction acts up the slope, 30 degrees above the horizontal, while the displacement of the mass is vertical, 4 times 2 equals 8 m upward. The work is the scalar product of the two: 5 times 8 times the cosine of the angle between them, which is 60 degrees, so the work is 40 times 0.5, that is 20 J.

Hinglish: Mass plane par slide nahi karta, isliye static friction mg sin 30 = 5 N up the slope. Friction ki direction 30 degree above horizontal, displacement 8 m vertical. Angle between them 60 degree, isliye work 5 x 8 x cos 60 = 20 J.

Q5 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A smooth inclined plane ends in a vertical circular loop of radius R. A small body is released from rest at height h on the incline. If the body exerts a force of three times its weight on the track at the highest point of the loop, then h = aR, where a is

Answer: By the third law the track pushes on the body with three times its weight at the top, so the normal reaction is 3mg downward. At the top the normal reaction and the weight both point toward the centre, so the centripetal equation is 3mg plus mg equals m v squared over R, giving v squared equal to 4gR. Energy conservation from height h to the top of the loop, where the height is 2R, gives mg h equals mg times 2R plus half m times 4gR, so h equals 4R and a is 4.

Hinglish: Top par track ka force weight ka teen guna, matlab N = 3mg. Top par normal aur weight dono centre ki taraf, isliye 3mg + mg = mv^2/R, matlab v^2 = 4gR. Energy se mgh = mg(2R) + 1/2 m(4gR), isliye h = 4R aur a = 4.

Q6 (JEE Main 2026, 6 Apr Shift 1, 4 marks)

A raindrop of mass 0.001 kg falls from rest through a height of 1000 m and reaches the ground at 5 m per second. The work done on it by the unknown resistive force is (take g = 10 m s^-2)

Answer: Gravity does positive work m g h, which is 0.001 times 10 times 1000, so 10 J. The kinetic energy gained is half of 0.001 times 25, which is 0.0125 J. The net work equals the change in kinetic energy, so the resistive work is 0.0125 minus 10, which is about minus 9.99 J, closest to minus 9.98 J. The sign is negative because the resistive force opposes the fall.

Hinglish: Gravity ka work mgh = 0.001 x 10 x 1000 = 10 J, aur kinetic energy 1/2 x 0.001 x 25 = 0.0125 J. Net work se resistive work 0.0125 - 10 = -9.9875 J, matlab lagbhag -9.98 J, negative kyunki resistive force motion ke against hai.

Q7 (NCERT Exemplar, 1 mark)

A man squatting on the ground gets up and stands. The force of reaction of the ground on the man during the process is (a) constant and equal to m g (b) constant and greater than m g (c) variable but always greater than m g (d) at first greater than m g and later equal to m g

Answer: While getting up the man's centre of mass accelerates upward at first and then decelerates to rest. The reaction must exceed the weight to give the initial upward acceleration, and then drop back to equal the weight when the man stands still. At the end the normal reaction equals m g. The reaction is therefore greater than m g at first and later equal to it, so option (d) is correct.

Hinglish: Uthte waqt pehle centre of mass upar accelerate hota hai, isliye reaction weight se zyada honi chahiye; phir rukne par reaction mg ke barabar ho jaati hai. Isliye option (d) sahi hai.

Q8 (NCERT Exemplar, 1 mark)

A bicycle skids to a stop over 10 m while the road pushes it with 200 N opposite to its motion. The work done by the bicycle on the road is (a) +2000 J (b) -200 J (c) zero (d) -20000 J

Answer: By the third law the cycle pushes on the road with 200 N in the direction of motion, but the road does not move, so the displacement of the point of application is zero. Work is force times displacement, therefore the work done by the cycle on the road is zero. The work done by the road on the cycle is not zero, because the cycle does move; that work is what removes the kinetic energy.

Hinglish: Third law se cycle road par 200 N force direction of motion mein lagata hai, lekin road ka displacement zero hai, isliye cycle dwara road par kiya work zero. Road dwara cycle par work non-zero hai, aur wahi kinetic energy kam karta hai.

Q9 (NCERT Exemplar, 1 mark)

A body falls freely under gravity alone in a vacuum. Which quantity remains constant during the fall? (a) Kinetic energy (b) Potential energy (c) Total mechanical energy (d) Total linear momentum

Answer: In the absence of air resistance only gravity acts, and gravity is conservative, so the total mechanical energy stays constant. The kinetic energy rises and the potential energy falls by equal amounts. The momentum changes because the velocity changes, and neither energy term is constant on its own. Option (c) is correct.

Hinglish: Vacuum mein sirf gravity lagti hai, jo conservative hai, isliye total mechanical energy constant rehti hai. Kinetic energy badhti hai aur potential energy utni hi ghatti hai. Momentum badalta hai, option (c) sahi.

Q10 (NCERT Exemplar, 1 mark)

During an inelastic collision between two bodies, which quantity always remains conserved? (a) Total kinetic energy (b) Total mechanical energy (c) Total linear momentum (d) Speed of each body

Answer: In any collision with no external force the total linear momentum is conserved, whether the collision is elastic or not. Kinetic energy is not conserved in an inelastic collision, and mechanical energy is lost to deformation, heat and sound. The speed of each body certainly changes. So option (c) is correct, and momentum conservation is the one statement that holds in every collision.

Hinglish: Collision mein external force na ho to total linear momentum hamesha conserved rehta hai, elastic ho ya inelastic. Inelastic mein kinetic energy conserve nahi hoti, mechanical energy loss hoti hai, aur speeds badalti hain. Isliye option (c) sahi hai.