A boy with mass of $20 k g$ is moving on a skateboard at 2 $m s^{- 1}$ . His speed increases to 4 $m s^{- 1}$ when an external force is acted upon him. Find the work done by the external force on the boy?

12 J

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120 J

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14 J

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140 J

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Solution

The correct option is B $120 J$
Initial velocity $v_{1}$ = $2 m / s$
Final velocity be $v_{2}$ = $4 m / s$
Mass of the boy $= 20 k g$ .

Work done by the external force = Change in K.E.
Change in K.E $= \frac{m v_{2}^{2}}{2} - \frac{m v_{1}^{2}}{2}$

$= \frac{20 \times (16 - 4)}{2}$

$= 10 \times 12 = 120 J$

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A boy with mass of 20 kg is moving on a skateboard at 2 ms−1. His speed increases to 4 ms−1 when an external force is acted upon him. Find the work done by the external force on the boy?

The correct option is B 120 J Initial velocity v1 = 2 m/s Final velocity be v2 = 4 m/s Mass of the boy =20 kg. Work done by the external force = Change in K.E. Change in K.E =mv222–mv212 =20×(16–4)2 =10×12=120 J

A boy with mass of $20 k g$ is moving on a skateboard at 2 $m s^{- 1}$ . His speed increases to 4 $m s^{- 1}$ when an external force is acted upon him. Find the work done by the external force on the boy?