A spring with a constant of $1 \frac{k g}{s^{2}}$ $1 (kg)/s^2$ 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 $3 \frac{m}{s}$ $3 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-05-27T08:30:45

The compression is $= 4.24 m$ $=4.24m$

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The spring constant is $k = 1 k g s^{- 2}$ $k=1kgs^-2$

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 2 \cdot {(3)}^{2} = 9 J$ $KE=1/2*2*(3)^2=9J$

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

$P E = 9 J$ $PE=9J$

So,

$\frac{1}{2} k x^{2} = 9$ $1/2kx^2=9$

$x^{2} = 2 \cdot \frac{9}{1} = 18 m^{2}$ $x^2=2*(9)/(1)=18m^2$

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

A spring with a constant of 1kgs21 (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 3ms3 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?