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 $7 k g$ $7 kg$ and speed of $5 \frac{m}{s}$ $5 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-07-15T05:06:03

The spring will compress by $= 6.61 m$ $=6.61m$

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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 $m = 7 k g$ $m=7kg$

The speed is $u = 5 m s^{- 1}$ $u=5ms^-1$

$K E = \frac{1}{2} \cdot 7 \cdot {(5)}^{2} = \frac{145}{2} J$ $KE=1/2*7*(5)^2=145/2J$

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

$P E = \frac{175}{2} J$ $PE=175/2J$

So,

$\frac{1}{2} k x^{2} = \frac{175}{2}$ $1/2kx^2=175/2$

$x^{2} = \frac{175}{4} = 43.75 m^{2}$ $x^2=(175)/(4)=43.75m^2$

$x = \sqrt{43.75} = 6.61 m$ $x=sqrt(43.75)=6.61m$

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 7kg7 kg and speed of 5ms 5 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?