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Solve the following :

Question:

A particle of mass $0.3 \mathrm{~kg}$ is subjected to a force $F=-\mathrm{kx}$ with $\mathrm{k}=15 \mathrm{~N} / \mathrm{m}$. What will be its initial acceleration if it is released from a point $x=20 \mathrm{~cm}$ ?

Solution:

$F=-k x=m a$

$=>a=\frac{-k x}{m}=\frac{-15 \times 20 \times 10^{-2}}{0.3}=-10 \mathrm{~m} / \mathrm{s}^{2}$

(-ve sign $=>^{a}$ is opposite to displacement)

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