Question

A spring with k = 53 N/m hangs vertically next to a ruler. The end of the spring is next to the 18 cm mark on the ruler. If a 2.4 kg 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 on the ruler marks when the mass is at its lowest position?

Answer 1

Answer:

1.07 m

Explanation:

x = Compression of the spring

k = Spring constant = 53 N/m

Initial length = 18 cm

P = Kinetic energy

K = Kinetic energy

At the lowest point of the mass the energy conservation is as follows

$P_{ig}+P_{is}+K_i=P_{fg}+P_{fs}+K_f\\\Rightarrow mgx+0+0=mgx+\frac{1}{2}kx^2\\\Rightarrow x=\frac{2mg}{k}\\\Rightarrow x=\frac{2\times 2.4\times 9.81}{53}\\\Rightarrow x=0.89\ m$

At its lowest position the mark on the ruler will be

$x_f=0.18+0.89\\\Rightarrow x_f=1.07\ m$

The spring line will end up at 1.07 m