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Question

A position dependent force F=(x23) N acts on a small body of mass 2 kg and displaces it from x=0 to x=5 m. The work done is

A
110 J
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B
803J
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C
952J
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D
Zero
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Solution

The correct option is B 803J
Work done by variable force is
W=x2x1F dxW=50(x23)dx=[x333x]50=125315W=803J
(b)
Why this question?

"The work done by a variable force is given by F.dr and in special case when both F and r are in x direction, it is given by W=F dx "

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