A carton lies on a plane tilted at an angle theta 29 degrees to the horizontal, with a friction of uk .12.If the carton starts from rest 7.90 up the plane from its base, what will be the carton's speed when it reaches the bottom of the incline?

Answer 1

The force of friction in the plane is
mu*mg*sinTheta
The gravity force down the plane is mgCosTheta

net force down the plane is
mgCosTheta-mu*mgSinTheta
the energy it gives to the box is that force times 7.9m
and that energy is equal to the KE at the bottom

(mgCosTheta-mu*mgSinTheta)*7.9=1/2 mv^2
solve for v.