You can push a 325 N trunk up a 20.0° inclined plane at a constant velocity by exerting a 209 N force parallel

You can push a 325 N trunk up a 20.0° inclined plane at a constant velocity by exerting a 209 N force parallel to the plane's surface. What force must be exerted on the trunk so that it would instead slide down the plane with a constant velocity?

we will use newton's second law

sum of forces = m a

if the trunk moves with constant velocity, a =0,

calling down the plane the positive direction, our forces are:

component of friction down the plane, friction down the plane (for the case where the trunk is pushed up the plane) and the pushing force of 209 N up the plane

newton's second law gives us

m g sin(theta) + u mg cos(theta) - 209N =0

we can use this to solve for u (coeff of friction):

325 sin 20 + u (325 cos 20) = 209

u = 0.32

in the second case where the trunk is sliding down, the frictional force acts up the plane to oppose motion, so we have now

mg sin 20 - u mg cos 20 - F =0 where F is the pushing force up the plane

325 sin 20 - 0.32 * 325 cos 20 = F

F = 13.4N