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?

Question

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?

Adrian Ayala · Accepted Answer

The spring will compress #2.12# metres.

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 #0# (because it's on the floor). It's kinetic energy will be given by #1/2mv^2#. Once it hits the spring, some of it's kinetic energy will be transferred to the spring. Therefore our equation will be #1/2mv_"initial"^2 = 1/2mv_"final"^2 + 1/2kx^2# We know that #v_"initial" = 3 m/s# and #v_"final = 0 m/s#. Therefore we get: #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!

Nathan Ashcroft · Answer

#=>x=sqrt(4.5) " m" approx 2.1 " m"#

Quantities given:

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

#=>E_k = 1/2mv^2#

The potential energy of a spring is given as:

#=>E_p = 1/2kx^2#

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.

#=>E_k = E_p#

#=>1/2mv^2 = 1/2kx^2#

#=>mv^2=kx^2#

We can solve for #x#, which is the displacement of the spring (assuming its initial displacement was #0#):

#=>x = sqrt((mv^2)/(k))#

Substituting known values:

#=>x=sqrt(((6)(3)^2)/(12))=sqrt(4.5) approx 2.1 " m"#

Carson Alexander · 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 given by:

$ KE_{initial} = \frac{1}{2}mv^2 $

where:
m = mass of the object (6 kg)
v = initial velocity of the object (3 m/s)

$ KE_{initial} = \frac{1}{2} * 6 \, kg * (3 \, m/s)^2 = 27 \, J $

When the object compresses the spring, all of its kinetic energy is converted into potential energy stored in the spring. The potential energy stored in a spring is given by:

$ PE_{spring} = \frac{1}{2}kx^2 $

where:
k = spring constant (12 (kg)/s^2)
x = compression of the spring (unknown)

Setting the initial kinetic energy equal to the potential energy stored in the spring:

$ KE_{initial} = PE_{spring} $

$ 27 \, J = \frac{1}{2} * 12 \, (kg)/s^2 * x^2 $

Solving for x:

$ x^2 = \frac{27 \, J}{\frac{1}{2} * 12 \, (kg)/s^2} $

$ x^2 = \frac{27 \, J}{6 \, (kg)/s^2} $

$ x^2 = 4.5 \, m^2 $

$ x = \sqrt{4.5 \, m^2} $

$ x \approx 2.12 \, m $

Therefore, the spring will compress approximately 2.12 meters.