A spring with a constant of $6 \frac{k g}{s^{2}}$ $6 (kg)/s^2$ 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 $7 \frac{m}{s}$ $7 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-05-09T11:28:10

The compression is $= 4.95 m$ $=4.95m$

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The spring constant is $k = 6 k g s^{- 2}$ $k=6kgs^-2$

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 3 \cdot {(7)}^{2} = 73.5 J$ $KE=1/2*3*(7)^2=73.5J$

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

$P E = 73.5 J$ $PE=73.5J$

So,

$\frac{1}{2} k x^{2} = 73.5$ $1/2kx^2=73.5$

$x^{2} = 73.5 \cdot \frac{2}{6} = 24.5 m^{2}$ $x^2=73.5*2/(6)=24.5m^2$

$x = \sqrt{24.5} = 4.95 m$ $x=sqrt(24.5)=4.95m$

A spring with a constant of 6kgs26 (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 7ms7 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?