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JEE Main 202225 JulyMorning Shift
Exam
JEE Main
Year
2022
Date
25 July
Shift
Morning Shift
Subject
Physics
Chapter
Work, Power and Energy

JEE Main 2022 · 25 July · Morning Shift · Physics · Work, Power and Energy

A body of mass 0 . 5kg travels on straight line path with velocity...

A body of mass 0.5kg0 . 5kg travels on straight line path with velocity v=(3x2+4)ms−1v = \left(3{x}^{2} + 4\right)m{s}^{- 1}. The net work done by the force during its displacement from x=0x = 0 to x=2mx = 2m is

Options

  1. A64J64J
  2. B60J60J
  3. C120J120J
  4. D128J128J

Correct answer

B. 60J60J

Step-by-step solution

Given: v=3x2+4v = 3{x}^{2} + 4

Initial and final velocities are vi=3(0)2+4=4ms−1{v}_{i} = 3{\left(0\right)}^{2} + 4 = 4m{s}^{- 1}

and vf=3(2)2+4=16ms−1{v}_{f} = 3{\left(2\right)}^{2} + 4 = 16m{s}^{- 1}

From work-energy theorem,

⇒\DeltaKE=12m(vf2−vi2)⇒ \DeltaKE = \dfrac{1}{2}m\left({v}_{f}^{2} - {v}_{i}^{2}\right)

⇒\DeltaKE=12×12(162−42)⇒ \DeltaKE = \dfrac{1}{2} \times \dfrac{1}{2}\left({16}^{2} - {4}^{2}\right)

⇒\DeltaKE=W=60J⇒ \DeltaKE = W = 60J

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