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 $20 cm$ away from orgin?

$(a)\;3\;m/s^2 \quad (b)\;15\; m/s^2\quad (c)\;5\;m/s^2 \quad (d)\;-10\;m/s^2$

1 Answer

$F=-kx=ma$
$a=\large\frac{-kx}{m}$
$a=\large\frac{-(15)(20 \times 10 ^{-2})}{0.3}$
$\quad=-10 m/s^2$
Hence d is the correct answer.

answered Jul 15, 2013 by
edited Feb 10, 2014 by meena.p

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