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Q.

A particle of mass 500 gm is moving in a straight line with velocity v=bx5/2. The work done by the net force during its displacement from x = 0 to x = 4 m is : (Take b=0.25 m−3/2s−1).

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a

2 J

b

4 J

c

16 J

d

8 J

answer is D.

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Detailed Solution

detailed_solution_thumbnail

By work energy theorem
Work done by all forces =Δ K.E. 
⇒W=12mvf2−12mvi2⇒W=12×0.5×(0.25)2×(4)5⇒W=16J 

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