A spring with a constant of #5 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #6 kg# and speed of #8 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 #5 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #6  kg# and speed of #8 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Emily Abbate · Accepted Answer

The compression is #=8.76m#

Mass of the object is #m=6kg#

Speed of the object is #v=8ms^-1#

The kinetic energy of the object is

#KE=1/2mv^2=1/2*6*8^2=192J#

This energy will be stored in the spring

#E=1/2kx^2#

The spring constant is #k=5kgs^-2#

Let the compression of the spring be #=xm#

Therefore,

#192=1/2*5*x^2#

#x^2=(2*192)/5=76.8m^2#

The compression is

#x=sqrt76.8=8.76m#

Leo Abeyta · Answer

undefined To find the compression of the spring, we can use the conservation of mechanical energy principle, equating the initial kinetic energy of the object to the potential energy stored in the compressed spring.

The initial kinetic energy of the object is given by $ \frac{1}{2}mv^2 $, where $ m $ is the mass of the object and $ v $ is its velocity. Plugging in the values, we get: 
$ KE = \frac{1}{2}(6 kg)(8 m/s)^2 = 192 J $.

The potential energy stored in the spring when compressed is given by $ \frac{1}{2}kx^2 $, where $ k $ is the spring constant and $ x $ is the compression distance. We can rearrange this equation to solve for $ x $:
$ PE = KE $
$ \frac{1}{2}kx^2 = 192 J $
$ x^2 = \frac{2 \times 192 J}{k} $
$ x^2 = \frac{384 J}{5 (kg)/s^2} $
$ x^2 = 76.8 m^2 $
$ x ≈ 8.75 m $

Therefore, the spring will compress approximately 8.75 meters.