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

Explanation:

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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