A spring with a constant of #6 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #7 kg# and speed of #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 #6 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #7  kg# and speed of #4 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

Julia Aguilar · Accepted Answer

The compression is #=4.32m#

The spring constant is #k=6kgs^-2#

The kinetic energy of the object is

#KE=1/2m u^2#

The mass is #m=7kg#

The speed is #u=4ms^-1#

#KE=1/2*7*(4)^2=56J#

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

#PE=56J#

The spring constant is #=6kgs^-2#

So,

#1/2kx^2=56#

#x^2=2*56/6=18.67m^2#

#x=sqrt(18.67)=4.32m#

William Audy · Answer

undefined To find the compression of the spring, we can use the conservation of mechanical energy. The kinetic energy of the object before the collision will be equal to the potential energy stored in the compressed spring after the collision.

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

Where:
$ m = 7 $ kg (mass of the object)
$ v = 4 $ m/s (speed of the object)

Substituting the values, we get:
$$ KE = \frac{1}{2}(7 \times 4^2) = 56 \, \text{J} $$

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

Where:
$ k = 6 \, \text{kg/s}^2 $ (spring constant)
$ x $ is the compression of the spring (in meters)

Equating the kinetic energy before the collision to the potential energy stored in the spring after the collision:
$$ KE = PE $$

$$ 56 = \frac{1}{2} \times 6 \times x^2 $$

Solving for $ x $:
$$ x = \sqrt{\frac{56 \times 2}{6}} \approx 3.29 \, \text{m} $$

So, the spring will compress approximately $ 3.29 $ meters.