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 $8 k g$ $8 kg$ and speed of $4 \frac{m}{s}$ $4 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2017-03-26T15:09:15

The spring will compress by $= 5.66 m$ $=5.66m$

Mass of the object is $m = 8 k g$ $m=8kg$

Speed of the object is $v = 4 m s^{- 1}$ $v=4ms^-1$

The kinetic energy of the object is

$K E = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 8 \cdot 4^{2} = 64 J$ $KE=1/2mv^2=1/2*8*4^2=64J$

This energy will be stored in the spring

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

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

Therefore,

$64 = \frac{1}{2} \cdot 4 \cdot x^{2}$ $64=1/2*4*x^2$

$x^{2} = \frac{64}{2} = 32 m^{2}$ $x^2=64/2=32m^2$

$x = \sqrt{32} = 5.66 m$ $x=sqrt32=5.66m$

A spring with a constant of 4 (kg)/s^24kgs24 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg8kg8 kg and speed of 4 m/s4ms4 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?