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Question
A body of mass 0.5 kg travels in a straight line with velocity v = `ax^(3/2)` where a = `5 m^(1/2)` s-1. What is the work done by the net force during its displacement from x = 0 to x = 2 m?
Solution
Here m = 0.5 kg
u=`a x^(3/2)`, a = `5 m^(-1/2)` s-1.
Initial velocity at x = 0, v1 = a x 0 = 0
Final velocity at x = 2, v2 = `a2^(3/2)` = `5 xx (2)^(3/2)`
Work done = increase in K.E
= `1/2` m(v22-v12)
= `1/2 xx 0.5[(5 xx 2^(3/2))^2 – 0]` = 50 J.
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