A spring with a constant of #2 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #4 kg# and speed of #1 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 #2 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #4  kg# and speed of #1  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Julia Aguilar · Accepted Answer

The spring will compress by #=1.4m#

Mass of the object is #m=4kg#

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

The kinetic energy of the object is

#KE=1/2mv^2=1/2*4*1^2=2J#

This energy will be stored in the spring

#E=1/2kx^2#

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

Let the compression of the spring be #=xm#

Therefore,

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

#x^2=(2*2)/2=2m^2#

The compression is

#x=sqrt2=1.4m#

William Audy · Answer

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

The initial kinetic energy of the object is given by:
$ KE = \frac{1}{2} mv^2 $

where:
$ m = 4 \, \text{kg} $ (mass of the object)
$ v = 1 \, \text{m/s} $ (speed of the object)

$ KE = \frac{1}{2} \times 4 \times (1)^2 = 2 \, \text{J} $

The potential energy stored in the compressed spring is given by:
$ PE = \frac{1}{2} kx^2 $

where:
$ k = 2 \, \text{kg/s}^2 $ (spring constant)
$ x $ is the compression of the spring (in meters)

$ PE = KE $
\( \frac{1}{2} \