A spring with a constant of #2/3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #1/5 kg# and speed of #1/4 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 #2/3 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #1/5  kg# and speed of #1/4  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Isabella Alvarez · Accepted Answer

The spring will compress by #=0.14m#

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

The kinetic energy of the object is

#KE=1/2m u^2#

#KE=1/2*1/5*(1/4)^2=1/160J#

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

#PE=1/160J#

So,

#1/2kx^2=1/160#

#x^2=1/(80k)=1/(2/3*80)=3/160#

#x=sqrt(3/160)=0.14m#

Nora Arzate · Answer

undefined To find the compression of the spring, you can use the equation for elastic potential energy:

$$ E_{\text{elastic}} = \frac{1}{2} k x^2 $$

where $ E_{\text{elastic}} $ is the elastic potential energy stored in the spring, $ k $ is the spring constant, and $ x $ is the compression of the spring.

You can also find the kinetic energy of the object before collision:

$$ E_{\text{kinetic}} = \frac{1}{2} m v^2 $$

where $ E_{\text{kinetic}} $ is the kinetic energy of the object, $ m $ is the mass of the object, and $ v $ is the speed of the object.

Since energy is conserved in this system, you can set the initial kinetic energy of the object equal to the elastic potential energy stored in the spring after compression:

$$ \frac{1}{2} m v^2 = \frac{1}{2} k x^2 $$

Substitute the given values:

$$ \frac{1}{2} \left( \frac{1}{5} \right) \left( \frac{1}{4} \right)^2 = \frac{1}{2} \left( \frac{2}{3} \right) x^2 $$

Solve for $ x $:

$$ x = \sqrt{\frac{\left( \frac{1}{2} \right) \left( \frac{1}{5} \right) \left( \frac{1}{4} \right)^2}{\left( \frac{1}{2} \right) \left( \frac{2}{3} \right)}} $$

$$ x = \sqrt{\frac{1}{5} \cdot \frac{1}{4} \cdot \frac{3}{2}} $$

$$ x = \sqrt{\frac{3}{40}} $$

$$ x = \frac{\sqrt{3}}{4} \approx 0.433 \, \text{m} $$