A spring with a constant of $5 \frac{k g}{s^{2}}$ $5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $5 k g$ $5 kg$ and speed of $8 \frac{m}{s}$ $8 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-10-16T06:28:34

The spring will compress by $= 8 m$ $=8m$

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The spring constant is $k = 5 k g s^{- 2}$ $k=5kgs^-2$

The kinetic energy of the object is

$K E = \frac{1}{2} m u^{2}$ $KE=1/2m u^2$

The mass is $m = 5 k g$ $m=5kg$

The speed is $u = 8 m s^{- 1}$ $u=8ms^-1$

$K E = \frac{1}{2} \cdot 5 \cdot {(8)}^{2} = 160 J$ $KE=1/2*5*(8)^2=160J$

This kinetic energy will be stored in the spring as potential energy.

$P E = 160 J$ $PE=160J$

The spring constant is $= 5 k g s^{- 2}$ $=5kgs^-2$

So,

$\frac{1}{2} k x^{2} = 160$ $1/2kx^2=160$

$x^{2} = 2 \cdot \frac{160}{5} = 64 m^{2}$ $x^2=2*(160)/(5)=64m^2$

$x = \sqrt{64} = 8 m$ $x=sqrt(64)=8m$

A spring with a constant of 5kgs25 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 5kg5 kg and speed of 8ms8 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?