A particle of mass $\sqrt{2}$ kg is moving in a circular path of radius 1 m. Speed of the particle is increasing at a rate $1 m / s^{2}$ . Then the force acting on the particle when its speed is $1 m / s$ is

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2 N

$2 \sqrt{2} N$

$\sqrt{2} N$

$4 \sqrt{2} N$

answer is A.

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Detailed Solution

Centripetal acceleration, $a_{n} = \frac{V^{2}}{r} = \frac{{(1)}^{2}}{1} \frac{m}{s^{2}} = 1 \frac{m}{s^{2}}$

Tangential acceleration, $a_{t} = 1 \frac{m}{s^{2}}$

Total acceleration, $a = \sqrt{a_{n}^{2} + a_{t}^{2}} = \sqrt{1^{2} + 1^{2}} \frac{m}{s^{2}} = \sqrt{2} \frac{m}{s^{2}}$

$∴$ Force, $F = m a = \sqrt{2} \times \sqrt{2} N = 2 N$

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