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Q. A 16 kg block moving on a frictionless horizontal surface with a velocity of 4 m/s compresses an ideal spring and comes to rest. If the force constant of the spring be 100 N/m, then the spring is compressed by

Jharkhand CECEJharkhand CECE 2010Mechanical Properties of Solids

Solution:

Kinetic energy of the block is
$K=\frac{1}{2}mv^2$
This kinetic energy is equal to the work done by the block before coming to rest. The work done in compressing the spring through a distance x from its normal length is
$W=\frac{1}{2}kx^2$
$\therefore $ $\frac{1}{2}mv^2=\frac{1}{2}kx^2\Rightarrow x=v\sqrt{\frac{m}{k}}$
Given, v = 4 m/s, m = 16 kg, k = 100 N/m
= $x=4\times\sqrt{\frac{16}{100}}=1.6 m$