A spring with a constant of $3 \frac{k g}{s^{2}}$ $3 (kg)/s^2$ 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 · 2017-05-15T18:29:42

The compression is $= 7.35 m$ $=7.35m$

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The spring constant is $k = 3 k g s^{- 2}$ $k=3kgs^-2$

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,

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

$x^{2} = 81 \cdot \frac{2}{3} = 54 m^{2}$ $x^2=81*2/(3)=54m^2$

$x = \sqrt{54} = 7.35 m$ $x=sqrt(54)=7.35m$

A spring with a constant of 3 (kg)/s^23kgs23 (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 9 m/s9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?