A spring with a constant of $4 (kg)/s^2$ is lying on the ground with one end attached to a wall. An object with a mass of $2 kg$ and speed of $5 m/s$ collides with and compresses the spring until it stops moving. How much will the spring compress?

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Answer 1 · 2016-12-18T15:17:07

Use conservation of energy to show that the compression of the spring will be 3.54 m.

Initially, the mass has kinetic energy equal to $1/2 mv^2$ = $1/2 (2)(5)^2$ = 25 J

During the collision, this kinetic energy is converted completely into potential energy in the compressed spring, given by $1/2kx^2$

So, Once the mass is stopped, all the initial KE is now PE:

25 J = $1/2 kx^2$

Since k = $4(kg)/s^2$ we can solve for $x$

25 = 2 $x^2$

$x^2$ = 12.5 $m^2$

$x$ = 3.54 $m$

A spring with a constant of 4 (kg)/s^24 (kg)/s^2 is lying on the ground with one end attached to a wall. An object with a mass of 2 kg2 kg and speed of 5 m/s5 m/s collides with and compresses the spring until it stops moving. How much will the spring compress?