Question

A massless spring with force constant ????=200N/m hangs from the ceiling. A 2.0-kg block is attached to the free end of the spring and released. If the block falls 17 cm before starting back upwards, how much work is done by friction during its descent?

Answer 1

-0.4454 Joules

Step-by-step explanation:

m = Mass of block = 2 kg

h = Height of extension = 17 cm = x

g = Acceleration due to gravity = 9.81 m/s²

Potential energy of the spring

`P=mgh\\\Rightarrow P=2* 9.81* 0.17\\\Rightarrow P=3.3354\ J`

The kinetic energy of the spring

`K=(1)/(2)mx^2\\\Rightarrow K=(1)/(2)* 200* 0.17^2\\\Rightarrow K=2.89\ J`

In this system as the potential and kinetic energy is conserved from work energy equivalence we get

`W=P-K\\\Rightarrow W=2.89-3.3354\\\Rightarrow W=-0.4454\ J`

The work done by friction is -0.4454 Joules