## Velocity, Force of Friction, and Total Energy in a System Question

1. The problem statement, all variables and given/known data
A 200 kg boulder falls from rest off a 120.0 m high cliff. If the boulder experiences a frictional force of 280 N[U] as it falls, what would be the velocity just before it hits the ground?

2. Relevant equations

kinetic energy = (1/2)mv2
gravitational potential energy = mgh
work = (net force)(Δd)

3. The attempt at a solution
Total force = mgh = 200kg*9.8N/kg*120m = 235200 J

Wfriction = FΔd
Wfriction = 280 N[U]*120 m
Wfriction = 33600 J

total energy = kinetic energy + work done by friction
kinetic energy = 235200 J - 33600 J
kinetic energy = 201600 J
201600 J = (1/2)mv2
v = 45 m/s

However, according to my teacher, this answer is incorrect.

 PhysOrg.com science news on PhysOrg.com >> Hong Kong launches first electric taxis>> Morocco to harness the wind in energy hunt>> Galaxy's Ring of Fire
 What does your notation "N[U]" represent? If, as I suspect, it is saying that the friction force is proportional to velocity (this is typical for air drag), then your treatment of the friction as constant is not appropriate here.

 Tags energy, friction, kinetic, potential, velocity