A spring with a constant of $7 \frac{k g}{s^{2}}$ $7 (kg)/(s^2)$ is lying on the ground with one end attached to a wall. An object with a mass of $4 k g$ $4 kg$ 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 · 2017-08-23T09:28:48

The compression is $= 1.15 m$ $=1.15m$

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The spring constant is $k = 4 k g s^{- 2}$ $k=4kgs^-2$

The kinetic energy of the object is

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

The mass is $m = 4 k g$ $m=4kg$

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

$K E = \frac{1}{2} \cdot 4 \cdot {(2)}^{2} = 8 J$ $KE=1/2*4*(2)^2=8J$

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

$P E = 8 J$ $PE=8J$

The spring constant is $= 7 k g s^{- 2}$ $=7kgs^-2$

So,

$\frac{1}{2} k x^{2} = 8$ $1/2kx^2=8$

$x^{2} = \frac{8 \cdot 2}{7} = 2.29 m^{2}$ $x^2=(8*2)/(7)=2.29m^2$

$x = \sqrt{2.29} = 1.51 m$ $x=sqrt(2.29)=1.51m$

A spring with a constant of 7kgs27 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 4kg4 kg and speed of 2ms 2 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?