A child of mass 2 kg is moving on a skateboard at the speed of $2 m s^{- 1}$ . After a force acts on him, his speed becomes $3 m s^{- 1}$ . Find the change in kinetic energy.

1 J

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10 J

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10 J

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12 J

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Solution

The correct option is B
10 J

Let initial speed of the child be $v_{1}$ and final speed be $v_{2}$
$v_{1} = 2 m s^{- 1}$
$v_{2} = 3 m s^{- 1}$
Initial kinetic energy of child = $\frac{1}{2} m {v_{1}}^{2} = \frac{1}{2} \times 2 \times 2^{2} = 4 J$

Final kinetic energy of child = $\frac{1}{2} m {v_{1}}^{2} = \frac{1}{2} \times 2 \times 3^{2} = 9 J$

Change in Kinetic Energy = $9 J - 4 J = 5 J$

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A child of mass 2 kg is moving on a skateboard at the speed of 2 ms−1. After a force acts on him, his speed becomes 3 ms−1. Find the change in kinetic energy.

The correct option is B 10 J Let initial speed of the child be v1 and final speed be v2 v1=2 ms−1 v2=3 ms−1 Initial kinetic energy of child = 12mv12 = 12×2×22 = 4 J Final kinetic energy of child = 12mv12 = 12×2×32 = 9 J Change in Kinetic Energy = 9 J − 4 J = 5 J

A child of mass 2 kg is moving on a skateboard at the speed of $2 m s^{- 1}$ . After a force acts on him, his speed becomes $3 m s^{- 1}$ . Find the change in kinetic energy.