A spring with a constant of $1 \frac{k g}{s^{2}}$ $1 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $2 k g$ $2 kg$ and speed of $4 \frac{m}{s}$ $4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-02-28T05:54:04

Here mass $, m = 2$ $, m = 2$ $k g$ $kg$
Velocity , $v = 4$ $v = 4$ $m s^{- 1}$ $ms^-1$
spring constant, $k = 1$ $k=1$ $k g s^{- 2}$ $kgs^-2$

By conservation of mechanical energy

PE of compressed spring= Initial kinetic energy of the moving object

$:.1/2*k*x^2=1/2*mv^2$ , where x = compression of spring
$=>x=sqrt((mv^2)/k)=sqrt(2*4^2/1)=4sqrt2m$

A spring with a constant of 1 (kg)/s^21kgs21 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2 kg2kg2 kg and speed of 4 m/s4ms4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?