Work Energy Power

JEE Main level, single correct, medium

36 of 183

Question 36 of 183

A body of mass 11 kg moves along a straight line with a velocity v=2x2v = 2x^2. The work done by the body during displacement from x=0x = 0 to 55 m is __________ J.

Show answer and solution

Answer: (C) 12501250

Given mass m=1m = 1 kg and velocity v=2x2v = 2x^2.

According to the work-energy theorem, the work done by the net force is equal to the change in kinetic energy of the body.

W=ΔK=Kf−KiW = \Delta K = K_f - K_i

At x=0x = 0 m, initial velocity vi=2(0)2=0v_i = 2(0)^2 = 0 m/s.

At x=5x = 5 m, final velocity vf=2(5)2=50v_f = 2(5)^2 = 50 m/s.

Initial kinetic energy Ki=12mvi2=0K_i = \dfrac{1}{2} m v_i^2 = 0 J.

Final kinetic energy Kf=12mvf2=12×1×(50)2=1250K_f = \dfrac{1}{2} m v_f^2 = \dfrac{1}{2} \times 1 \times (50)^2 = 1250 J.

Work done W=1250−0=1250W = 1250 - 0 = 1250 J.

Answer: 12501250

JEE Main question paper PDF

Open a shift to read the full paper on this page, or download it.

JEE Main 2025 Apr 2 Shift 1 question paper

Open Download PDF

Opening paper…

Explain it as a whiteboard videoStep-by-step, in Hinglish or English. Your first video needs no login.Solved it on paper but got it wrong?Upload a photo of your working. Mistake Doctor marks the exact line that broke.
See it move: Collisions, centre of mass and rockets simulationDrag the sliders and watch the numbers in this question change.

More Physics practice

JEE Main Physics practice question on Work Energy Power (Energy), with the answer and a step-by-step solution. Scan a question from your own book or paper to get a whiteboard video.