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14 October, 12:34

A force F of magnitude 2x^3 is applied to stop a particle moving with an initial velocity of v0. The particle travels from x=0 to x = D in coming to a stop after the force F is applied. The work done by F is:a) 1/2 D^3b) - D^4c) D^4d) 2D^3e) - D^3f) - 1/2 D^4g) 1/2D^4h) - 1/2D^3

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  1. 14 October, 16:31
    0
    Given that

    F=2x³

    Work is given as

    The range of x is from x=0 to x=D

    W=-∫f (x) dx

    Then,

    W=-∫2x³dx from x=0 to x=D

    W = - 2x⁴/4 from x=0 to x=D

    W=-2 (D⁴/4-0/4)

    W=-D⁴/2

    W=1/2D⁴

    The correct answer is F
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