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