Question

A 5 kg brick is dropped from a height of 12m on a spring with a spring constant 8 kN/m. If the spring has unstretched length of 0.5m, find: a) The shortest length the spring will be compressed before recoil, and
b) The final length of the spring once the whole system become static.

Answer 1

Answer:

0.1164 m

0.49387 m

Explanation:

m = Mass = 5 kg

g = Acceleration due to gravity = 9.81 m/s²

h = Height from which brick falls = 12 m

k = Spring constant = 8 kN/m

a) Potential energy of the brick

$PE=mgh\\\Rightarrow PE=5\times 9.81\times 12\\\Rightarrow PE=588.6\ J$

Potential energy in spring

$PE=\frac{1}{2}\times kx^2\\\Rightarrow x=\sqrt{\frac{PE\times 2}{k}}\\\Rightarrow x=\sqrt{\frac{588.6\times 2}{8000}}\\\Rightarrow x=0.3836\ m$

The compression of spring = 0.5-0.3836 = 0.1164 m

b) Weight of the brick

$F=5\times 9.81\\\Rightarrow F=49.05\ N$

$F=kx\\\Rightarrow x=\frac{F}{k}\\\Rightarrow x=\frac{49.05}{8000}\\\Rightarrow x=0.00613\ m$

The final length of the spring = 0.5-0.00613 = 0.49387 m