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

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Answer 1 · 2017-04-11T16:49:35

The spring will compress by $= 10.58 m$ $=10.58m$

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

The kinetic energy of the object is

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

$K E = \frac{1}{2} \cdot 7 \cdot 8^{2} = 224 J$ $KE=1/2*7*8^2=224J$

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

$P E = 224 J$ $PE=224J$

So,

$\frac{1}{2} k x^{2} = 224$ $1/2kx^2=224$

$x^{2} = \frac{448}{k} = \frac{448}{4} = 112$ $x^2=448/k=448/4=112$

$x = \sqrt{112} = 10.58 m$ $x=sqrt112=10.58m$

A spring with a constant of 4 (kg)/(s^2)4kgs24 (kg)/(s^2) is lying on the ground with one end attached to a wall. An object with a mass of 7 kg 7kg7 kg and speed of 8 m/s8ms 8 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?