Work/force calculation problem?

Question

A worker removes a 5kg carton of soft drink from the top of a stack in a warehouse by sliding it down a ramp. The ramp has a 15 degree incline to the horizontal. After sliding 12m down the rap, the carton has a speed of 6m/s.

How much work does the carton do to overcome ramp friction?

What is the average frictional force the carton experiences?

Answer 1

The change in elevation is ∆h = L*sinθ  where L is the length of the ramp and θ the angle.  The potential energy change is then m*g*∆h = m*g*L*sinθ.  The kinetic energy change is 0.5*m*v²  The difference is the work done by the carton to overcome friction:

Ef = m*g*L*sinθ. - 0.5*m*v²

The work done by friction is also Fr*L, so the average force of friction is

Fr = m*g*sinθ. - 0.5*m*v²/L

Ef = 62.2 J

Fr = 5.18 N