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 $3 k g$ $3 kg$ and speed of $12 \frac{m}{s}$ $12 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-21T12:09:21

The compression is $= 9.3 m$ $=9.3m$

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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 $m = 3 k g$ $m=3kg$

The speed is $u = 12 m s^{- 1}$ $u=12ms^-1$

$K E = \frac{1}{2} \cdot 3 \cdot {(12)}^{2} = 216 J$ $KE=1/2*3*(12)^2=216J$

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

$P E = 216 J$ $PE=216J$

So,

$\frac{1}{2} k x^{2} = 216$ $1/2kx^2=216$

$x^{2} = \frac{2 \cdot 216}{5} = 86.4 m^{2}$ $x^2=(2*216)/(5)=86.4m^2$

$x = \sqrt{86.4} = 9.3 m$ $x=sqrt(86.4)=9.3m$

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 3kg3 kg and speed of 12ms12 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?