A particle of mass $0.3 k g$ 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 $20 c m$ away from the origin?

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Solution

Given:
The mass of the particle is $0.3 k g$
The value of the constant $k$ is $15 N / m$ .

Since the force acting on the particle is :
$F = - k x$

The point of action of the force is $20 c m$ away from the origin So,
$| F | = k x \Rightarrow 15 \times 20 \times 10^{- 2}$

$| F | = 3 N$

Now the acceleration of the particle s given as:
$a = \frac{| F |}{m}$

$a = \frac{3}{0.3} = 10 m / s^{2}$

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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 \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?

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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 20cm away from the origin?

Given:The mass of the particle is 0.3 kgThe value of the constant k is 15 N/m.Since the force acting on the particle is :F=−kxThe point of action of the force is 20cm away from the origin So,|F|=kx⇒15×20×10−2|F|=3 NNow the acceleration of the particle s given as:a=|F|ma=30.3=10 m/s2

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