A spring with a constant of $\frac{1}{5} \frac{k g}{s^{2}}$ $1/5 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $1 k g$ $1 kg$ and speed of $2 \frac{m}{s}$ $2 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-06-17T12:06:46

The compression is $= 4.47 m$ $=4.47m$

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The spring constant is $k = \frac{1}{5} k g s^{- 2}$ $k=1/5kgs^-2$

The kinetic energy of the object is

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

The mass is $m = 1 k g$ $m=1kg$

The speed is $u = 2 m s^{- 1}$ $u=2ms^-1$

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

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

$P E = 2 J$ $PE=2J$

So,

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

$x^{2} = \frac{4}{\frac{1}{5}} = 20 m^{2}$ $x^2=(4)/(1/5)=20m^2$

$x = \sqrt{20} = 4.47 m$ $x=sqrt(20)=4.47m$

A spring with a constant of 1/5 (kg)/s^215kgs21/5 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 1 kg1kg1 kg and speed of 2 m/s2ms2 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?