A particle of mass $0.3 \mathrm{~kg}$ is subjected to a force $F=-k x$ with $k=15 \mathrm{~N} / \mathrm{m}$.…
A particle of mass $0.3 \mathrm{~kg}$ is subjected to a force $F=-k x$ with $k=15 \mathrm{~N} / \mathrm{m}$. What will be its initial acceleration if it is released from a point $20 \mathrm{~cm}$ away from the origin?
$3 \mathrm{~m} / \mathrm{s}^2$
$15 \mathrm{~m} / \mathrm{s}^2$
$5 \mathrm{~m} / \mathrm{s}^2$
$10 \mathrm{~m} / \mathrm{s}^2$
Solution
$a=\frac{k x}{m}=10 \mathrm{~m} / \mathrm{s}^2$
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The force acting on the particle is F = -kx.
By Newton's second law, F = ma,
Firstly, we need to convert the distance from cm to m, so x = 20 cm = 0.2 m.
Substituting F = -kx into F = ma gives -kx = ma.
We can rearrange this to find the acceleration: a = -kx/m.
Substituting in the given values gives a = -(15 N/m * 0.2 m) / 0.3 kg = -10 m/s^2.
The negative sign indicates that the acceleration is in the opposite direction
to the displacement from the origin, as expected for a restoring force.
Therefore,
the initial acceleration of the particle is 10 m/s² in the direction towards the origin