A spring with a constant of $4 \frac{k g}{s^{2}}$ $4 (kg)/s^2$ 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 $9 \frac{m}{s}$ $9 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

1373 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-29T14:14:12

The compression is $= 9 m$ $=9m$

enter image source here

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 $u = 9 m s^{- 1}$ $u=9ms^-1$

$K E = \frac{1}{2} \cdot 4 \cdot {(9)}^{2} = 162 J$ $KE=1/2*4*(9)^2=162J$

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

$P E = 162 J$ $PE=162J$

So,

$\frac{1}{2} k x^{2} = 162$ $1/2kx^2=162$

$x^{2} = \frac{162 \cdot 2}{4} = 81 m^{2}$ $x^2=(162*2)/(4)=81m^2$

$x = \sqrt{81} = 9 m$ $x=sqrt(81)=9m$

A spring with a constant of 4kgs24 (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 9ms9 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?