A spring with k = 55 N/m hangs vertically next to a ruler. The end of the spring is next to the 18-cm mark on the ruler.

Question

A spring with kk = 55 N/mN/m hangs vertically next to a ruler. The end of the spring is next to the 18-cmcm mark on the ruler. If a 3.0-kgkg mass is now attached to the end of the spring, and the mass is allowed to fall, where will the end of the spring line up with the ruler marks when the mass is at its lowest position?

Express your answer to two significant figures and include the appropriate units.

Answer 1

When we attach the mass the spring will stretch until the point where the mass momentarily stops before traveling back up again. If we let the additional extension of the spring be x, then the gravitational energy of the mass relative to the bottom position will be equal to the spring energy at the bottom.

We then get mgx = 1/2 * k * x^2

this allows us to simplify to find x = 2mg/k

Plugging in m = 3, k =55 and g = 9.81 we get x = 1.07 meters = 107 cms

Since we started at 18 cms, the new ruler marking would be 18 + 107 = 125 cms

As we can only do 2 significant figures this becomes 130 cms

(Note if we use 9.8 instead of 9.81, the answer is just below 125 cms so with two sig figs it would be 120 cms)