Question:

A body of mass 0.5 kg travels in a straight line with velocity v =a x 3/2^{3/2} where a = 5 m1/2^{-1/2} s–1 . What is the work done by the net force during its displacement from x = 0 to x = 2 m ?

Updated On: Nov 3, 2023
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Solution and Explanation

Mass of the body, m = 0.5 kg 
Velocity of the body is governed by the equation, v = a x3/2^{3/2} with a = 5 m 1/2^{-1/2} s - 1
Initial velocity, u (at x = 0) = 0 
Final velocity v (at x = 2 m) = 102m/s10\sqrt{2}{m/s}
Work done, W = Change in kinetic energy = 12\frac{1}{2} m(v2 - u2)
12\frac{1}{2}× [(102)2(10\sqrt{2})^2 - (0)2^2]
12\frac{1}{2} × 0.5 × 10 × 10 × 2
= 50 J
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