A spring with a constant of #1# #kgs^-2# is lying on the ground with one end attached to a wall. An object with a mass of #8# #kg# and speed of #3# #ms^-1# collides with and compresses the spring until it stops moving. How much will the spring compress?

Question

A spring with a constant of #1# #kgs^-2# is lying on the ground with one end attached to a wall. An object with a mass of #8# #kg# and speed of #3#  #ms^-1# collides with and compresses the spring until it stops moving. How much will the spring compress?

Benjamin Asiedu · Accepted Answer

In this instance kinetic energy is converted into spring potential energy. The spring compresses by #8.5# #m#.

The moving mass's initial kinetic energy can be found using: #E_k=1/2mv^2=1/2*8*3^2=36# #J# The following equation predicts that all of this energy will be transformed into spring potential energy in the spring, assuming no friction: #E_p=1/2kx^2# where #k# is the spring constant and #x# is the distance compressed (or expanded). We can rearrange this to make #x# the subject: #x=sqrt((2E_p)/k) = sqrt((2*36)/1) = sqrt 72 =8.5# #m#

Nora Arzate · Answer

undefined Use the conservation of mechanical energy:

$$ \frac{1}{2}mv^2 = \frac{1}{2}kx^2 $$

where $ m = 8 \, \text{kg} $ is the mass, $ v = 3 \, \text{m/s} $ is the speed, and $ k = 1 \, \text{kg/s}^2 $ is the spring constant. Solve for $ x $.