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

0.1164 m

0.49387 m

Step-by-step 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* 9.81* 12\\\Rightarrow PE=588.6\ J`

Potential energy in spring

`PE=(1)/(2)* kx^2\\\Rightarrow x=\sqrt{(PE* 2)/(k)}\\\Rightarrow x=\sqrt{(588.6* 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* 9.81\\\Rightarrow F=49.05\ N`

`F=kx\\\Rightarrow x=(F)/(k)\\\Rightarrow x=(49.05)/(8000)\\\Rightarrow x=0.00613\ m`

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