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A force of magnitude 10 N acts on a body of 4 kg moving initially in the direction of the force with a speed of $5 m/s$ . The distance through which the force must act in order to increase the speed of the body to $10m/s$ is

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25 m

5 m

`10 m

15 m

answer is C.

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Detailed Solution

F = 10 N,

$u = 5 m/s$ , m = 4 kg

$v = 5 m/s$

According to work energy theorem

$W = Δ K.E = \frac{1}{2} {mv}^{2} - \frac{1}{2} {mu}^{2}$

$F.S = \frac{1}{2} m (v^{2} - u^{2})$

$∴ S = \frac{\frac{1}{2} m (v^{2} - u^{2})}{F} = \frac{\frac{1}{2} (4) (10^{2} - 5^{2})}{10}$

$= \frac{100 - 25}{5} = \frac{75}{5} = 15 m$

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