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 $1 k g$ $1 kg$ and speed of $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-16T13:10:45

The compression is $= 4.02 m$ $=4.02m$

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

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 1 \cdot {(9)}^{2} = \frac{81}{2} J$ $KE=1/2*1*(9)^2=81/2J$

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

$P E = \frac{81}{2} J$ $PE=81/2J$

So,

$\frac{1}{2} k x^{2} = \frac{81}{2}$ $1/2kx^2=81/2$

$x^{2} = \frac{81}{5} = 16.2 m^{2}$ $x^2=(81)/(5)=16.2m^2$

$x = \sqrt{16.2} = 4.02 m$ $x=sqrt(16.2)=4.02m$

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 1kg1 kg and speed of 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?