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 #5 kg # and speed of # 7 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 #5 kg # and speed of # 7  m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Layla Ahmed · Accepted Answer

The distance is #=7m#

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

The kinetic energy of the object is

#KE=1/2m u^2#

#KE=1/2*5*(7)^2=122.5J#

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

#PE=122.5J#

So,

#1/2kx^2=122.5#

#x^2=2*(122.5)/(5)=49m^2#

#x=sqrt(49)=7m#

Elena Albarran · 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.

Given:
- $k = 5 \frac{\text{kg}}{\text{s}^2}$
- $m = 5 \text{ kg}$
- $v = 7 \text{ m/s}$

First, calculate the kinetic energy of the object using:

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

Then, equate $E_{\text{kinetic}}$ to $E_{\text{elastic}}$ and solve for $x$.