A spring with a constant of 2 (kg)/(s^2)2kgs2 is lying on the ground with one end attached to a wall. An object with a mass of 8 kg 8kg and speed of 5 m/s5ms collides with and compresses the spring until it stops moving. How much will the spring compress?

1 Answer
Apr 19, 2016

10m10m

Explanation:

Energy is conserved in this reaction.

Kinetic energy of the object is given by 1/2mv^212mv2, where mm is mass and vv is velocity, and potential energy stored in the spring is given by 1/2kx^212kx2, where kk is the spring constant and xx is the distance it compresses. If there is always the same amount of energy, then

1/2mv^2 = 1/2kx^212mv2=12kx2

therefore mv^2=kx^2

Rearranging this to make x, the distance of compression, the subject (because the question asks for x),

mv^2 = kx^2

(mv^2)/k = x^2

sqrt((mv^2)/k) = x.

Putting in values the question gives us, k=2(kg)/s^2, m=8kg, v=5m/s, then

x = sqrt((8*5^2)/2)

x=sqrt(200/2)

x=10m