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 $5 k g$ $5 kg$ 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?

Question

1989 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-07-11T07:13:58

The spring will compress by $= 5.9 m$ $=5.9m$

enter image source here

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

The kinetic energy of the object is

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

The mass is $m = 5 k g$ $m=5kg$

The speed is $u = 7 m s^{- 1}$ $u=7ms^-1$

$K E = \frac{1}{2} \cdot 5 \cdot {(7)}^{2} = 122.5 J$ $KE=1/2*5*(7)^2=122.5J$

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

$P E = 122.5 J$ $PE=122.5J$

So,

$\frac{1}{2} k x^{2} = 122.5$ $1/2kx^2=122.5$

$x^{2} = \frac{2 \cdot 122.5}{7} = 35 m^{2}$ $x^2=(2*122.5)/(7)=35m^2$

$x = \sqrt{35} = 5.9 m$ $x=sqrt(35)=5.9m$

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 5kg5 kg and speed of 7ms 7 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?