# A spring with a constant of #12 (kg)/s^2# is lying on the ground with one end attached to a wall. An object with a mass of #6 kg# and speed of #3 m/s# collides with and compresses the spring until it stops moving. How much will the spring compress?

##### 2 Answers

The spring will compress

#### Explanation:

We can either do this using conservation of energy or kinematics. I'll use conservation of energy.

The object's potential energy will be at

#1/2mv_"initial"^2 = 1/2mv_"final"^2 + 1/2kx^2#

We know that

#1/2(6)(3)^2 = 1/2(6)(0)^2 + 1/2(12)x^2#

#27 = 6x^2#

#x = sqrt(27/6) ~~2.12 m#

Hopefully this helps!

#### Explanation:

Quantities given:

#k = 12 " kg" * "m/s"# #m = 6 " kg"# #v = 3 " m/s"#

We can calculate the kinetic energy of the object at the moment before it contacts the spring:

The potential energy of a spring is given as:

When the object comes to rest, it will be because the object has lost all of its kinetic energy and the energy has been transferred into the spring.

We can solve for

Substituting known values: