A spring with a constant of #3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #3 kg# and speed of #9 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

A spring with a constant of #3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #3  kg# and speed of #9  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Julia Aguilar · Accepted Answer

The compression is #=9m#

The spring constant is #k=1kgs^-2#

The kinetic energy of the object is

#KE=1/2m u^2#

The mass is #m=3kg#

The speed is #u=9ms^-1#

#KE=1/2*3*(9)^2=243/2J#

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

#PE=243/2J#

The spring constant is #=3kgs^-2#

So,

#1/2kx^2=243/2#

#x^2=(243)/(3)=81m^2#

#x=sqrt(81)=9m#

Lily Adams · Answer

undefined To find the compression of the spring, use the conservation of mechanical energy equation, $ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 $, where $ m = 3 \, \text{kg} $, $ v = 9 \, \text{m/s} $, and $ k = 3 \, \text{kg/s}^2 $. Solving for $ x $, you get $ x = \sqrt{\frac{mv^2}{k}} $. Plugging in the values, $ x = \sqrt{\frac{(3 \, \text{kg})(9 \, \text{m/s})^2}{3 \, \text{kg/s}^2}} = 9 \, \text{m} $. Therefore, the spring will compress by $ 9 \, \text{m} $.