A spring with a constant of $9 \frac{k g}{s^{2}}$ $9 (kg)/s^2$ 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 $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-11T11:27:05

The height is $= 4.24 m$ $=4.24m$

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The spring constant is $k = 9 k g s^{- 2}$ $k=9kgs^-2$

The kinetic energy of the object is

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

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

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

$P E = 81 J$ $PE=81J$

So,

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

$x^{2} = 2 \cdot \frac{81}{9} = 18 m^{2}$ $x^2=2*(81)/(9)=18m^2$

$x = \sqrt{18} = 4.24 m$ $x=sqrt(18)=4.24m$

A spring with a constant of 9kgs29 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2kg2 kg and speed of 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?