A spring with a constant of $6$ $6$ $k g s^{- 2}$ $kgs^-2$ is lying on the ground with one end attached to a wall. An object with a mass of $7$ $7$ $k g$ $kg$ and speed of $2$ $2$ $m s^{- 1}$ $ms^-1$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-04-28T10:12:57

When the initial kinetic energy, $14$ $14$ $J$ $J$ is converted to spring potential energy, the spring will be compressed by $2.16$ $2.16$ $m$ $m$ .

The kinetic energy of the moving mass will be converted to spring potential energy.

Initial kinetic energy:

$E_{k} = \frac{1}{2} m v^{2} = \frac{1}{2} \cdot 7 \cdot 2^{2} = 14$ $E_k=1/2mv^2=1/2*7*2^2=14$ $J$ $J$ .

Final spring potential energy:

$E_{p} = \frac{1}{2} k x^{2}$ $E_p=1/2kx^2$ .

Rearranging: $x = \sqrt{\frac{2 E_{p}}{k}} = \sqrt{\frac{2 \cdot 14}{6}} \approx 2.16$ $x=sqrt((2E_p)/k)=sqrt((2*14)/6)~~2.16$ $m$ $m$

A spring with a constant of 666 kgs^-2kgs−2kgs^-2 is lying on the ground with one end attached to a wall. An object with a mass of 777 kgkg kg and speed of 222 ms^-1ms−1ms^-1 collides with and compresses the spring until it stops moving. How much will the spring compress?