# Kinetic energy of a crate

1. Mar 14, 2009

### ramenmeal

1. The problem statement, all variables and given/known data

A crate of mass 10.2 kg is pulled up a rough incline with an initial speed of 1.42 m/s. The pulling force is 94.0 N parallel to the incline, which makes an angle of 20.7° with the horizontal. The coefficient of kinetic friction is 0.400, and the crate is pulled 4.92 m.

(b) Determine the increase in internal energy of the crate-incline system due friction.
(d) What is the change in kinetic energy of the crate?
(e) What is the speed of the crate after being pulled 4.92 m?

2. Relevant equations

KE = .5mv^2
frictional force = u(Fn)
work = F x distance

3. The attempt at a solution

for b, i thought it should just be the force of friction x distance so i did uFn(4.92)
for the others i dont really know

2. Mar 14, 2009

### BAnders1

The total energy of the system is conserved, that is, $$E=K+U+E_{int}=constant$$.
In general, $$K=1/2mv^2, U=mgy$$ and $$E_{int}$$ is the energy lost to friction.

Draw a picture of the initial and final states of the system, and keep in mind that the energy is the same in both states, though $$K, U$$, and $$E_{int}$$ may have different values.