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 #2 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 #2  kg# and speed of #9  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Amelia Adams · Accepted Answer

The compression is #=7.35m#

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

The kinetic energy of the object is

#KE=1/2m u^2#

#KE=1/2*2*(9)^2=81J#

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

#PE=81J#

So,

#1/2kx^2=81#

#x^2=81*2/(3)=54m^2#

#x=sqrt(54)=7.35m#

Nathan Ashcroft · Answer

undefined To find the compression of the spring, we can use the principle of 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: 
$KE_{initial} = \frac{1}{2}mv^2$
$KE_{initial} = \frac{1}{2}(2 \, kg)(9 \, m/s)^2$
$KE_{initial} = 81 \, J$

The potential energy stored in the spring is given by:
$PE_{spring} = \frac{1}{2}kx^2$
Where $k$ is the spring constant and $x$ is the compression distance.

Setting the initial kinetic energy equal to the potential energy stored in the spring:
$KE_{initial} = PE_{spring}$
$81 \, J = \frac{1}{2}(3 \, kg/s^2)x^2$

Solving for $x$:
$x^2 = \frac{2 \times 81}{3}$
$x^2 = 54$
$x \approx 7.35 \, m$

Therefore, the spring will compress approximately $7.35 \, m$.