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
Mar 26, 2018

Answer:

The spring will compress #2.12# metres.

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 #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!

Mar 26, 2018

Answer:

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

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:

#=>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"#