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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Solution

Force $F = - k x = - 15 x$
At $x = 20 c m = 0.2 m$ , $F = - 15 \times 0.2 = - 3$
Acceleration $a = \frac{F}{m} = \frac{- 3}{0.3} = - 10 m / s^{2}$

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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 ?

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?

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?

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:

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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?

Force F=−kx=−15xAt x=20 cm=0.2 m, F=−15×0.2=−3Acceleration a=Fm=−30.3=−10 m/s2

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?

Force $F = - k x = - 15 x$
At $x = 20 c m = 0.2 m$ , $F = - 15 \times 0.2 = - 3$
Acceleration $a = \frac{F}{m} = \frac{- 3}{0.3} = - 10 m / s^{2}$