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Net work done when velocity is given as a function of position (work-energy theorem)

This type of question, asked 4 times: 29 Jun 2022, 25 Jul 2022, 9 Apr 2024, 6 Apr 2026 S2. All 4 questions below, oldest first.

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  1. Question 1

    29 Jun 2022

    8
    A particle of mass 500gm500gm is moving in a straight line with velocity v=bx5/2v=bx^{5/2}. The work done by the net force during its displacement from x=0x = 0 to x=4 mx = 4\text{ }m is: (Take  b=0.25 m−3/2 s−1)\left. \ b = 0.25{\text{ }m}^{- 3/2}{\text{ }s}^{- 1} \right).

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  2. Question 2

    25 Jul 2022

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

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  3. Question 3

    9 Apr 2024

    14
    A particle of mass mm moves on a straight line with its velocity increasing with distance according to the equation v=αxv = \alpha\sqrt{x}, where α\alpha is a constant. The total work done by all the forces applied on the particle during its displacement from x=0x = 0 to x=dx = d, will be:

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  4. Question 4

    6 Apr 2026 S2

    4
    A body of mass 1 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 5 m is ____\_\_\_\_ J.

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