A particle of mass $0.3 k g$ is subjected to a force $F = k x$ with $k = 15 N m^{- 1}$ . What will be its initial acceleration if it is released from a point $20 c m$ away from the origin?

3 m s^{- 2}

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15 m s^{- 2}

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5 m s^{- 2}

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10 m s^{- 2}

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Solution

The correct option is D $10 m s^{- 2}$
$F = k x$
Let $a$ be the acceleration of particle when it release from $x = 0.2 m$
So, $m a = k x \Rightarrow a = \frac{k x}{m} = \frac{15 \times 0.2}{0.3} = 10 m / s^{2}$

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Particle of mass 0.3 kg is subjected to a force F = –kx with k = 15 N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin:

Q. A particle of mass

0.3 k g

is subjected to a force

F = - k x

with

k = 15 \frac{N}{m}

. What will be the magnitude of its initial acceleration (in

\frac{m}{s^{2}}

), if it is released from a point

x = 20 c m

from its equilibrium position?

Q. A particle of mass 0.3 kg is subjected to a force

F = - k x

with k = 15 N/m. What will be its initial acceleration if it is released from a point x = 20 cm?

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A particle of mass 0.3kg is subjected to a force F=kx with k=15 Nm−1. What will be its initial acceleration if it is released from a point 20cm away from the origin?

The correct option is D 10 ms−2F=kxLet a be the acceleration of particle when it release from x=0.2 mSo, ma=kx⇒a=kxm=15×0.20.3=10 m/s2

A particle of mass $0.3 k g$ is subjected to a force $F = k x$ with $k = 15 N m^{- 1}$ . What will be its initial acceleration if it is released from a point $20 c m$ away from the origin?

The correct option is D $10 m s^{- 2}$
$F = k x$
Let $a$ be the acceleration of particle when it release from $x = 0.2 m$
So, $m a = k x \Rightarrow a = \frac{k x}{m} = \frac{15 \times 0.2}{0.3} = 10 m / s^{2}$