A particle of mass 0.3 kg is subjected to a force $F = - k x$ with $k = 15 \frac{N}{m}$ . What will be the magnitude of its initial acceleration ( $i n \frac{m}{s^{2}}$ ) if it is released from a point x=20 cm from its equilibrium position?

Open in App

Solution

Assuming the direction of compression as negative,

$F = - k x = m a \Rightarrow a = \frac{- 15 (0.2)}{0.3} = - 10 m / s^{2}$

Suggest Corrections

Similar questions

Q. A 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 x = 20 cm?

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 \frac{N}{m}

. What will be the magnitude of its initial acceleration (

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

) if it is released from a point x=20 cm from its equilibrium position?

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:

Join BYJU'S Learning Program

A particle of mass 0.3 kg is subjected to a force F=−kx with k=15Nm. What will be the magnitude of its initial acceleration (inms2) if it is released from a point x=20 cm from its equilibrium position?

Assuming the direction of compression as negative, F=−kx=ma⇒a=−15(0.2)0.3=−10m/s2

A particle of mass 0.3 kg is subjected to a force $F = - k x$ with $k = 15 \frac{N}{m}$ . What will be the magnitude of its initial acceleration ( $i n \frac{m}{s^{2}}$ ) if it is released from a point x=20 cm from its equilibrium position?

Assuming the direction of compression as negative,

$F = - k x = m a \Rightarrow a = \frac{- 15 (0.2)}{0.3} = - 10 m / s^{2}$