Friction problem, please help!?

The force that is acting to accelerate the crate 'down the hill' on the plane is the parallel component of its weight force, F(par) = mg sin (theta) = 4.9m N.

We don't know the mass of the crate but this should cancel out as we go through and make it possible to complete the calculation.

The frictional force will be given by F(fric) = mu(k)*F(norm), where F(norm) = mg cos (theta).

Given that (theta) = 30 degrees, g = 9.8 and mu(k) = 0.16, the frictional force will have a magnitude of 1.36m N. This force will always oppose the direction of motion, so will act 'down the hill' as the crate slides up, then 'up the hill' as it slides down.

To find the time taken for the crate to reach the top of its motion, remember that u = 3.3 ms and v = 0. The acceleration will be given by a=F/m for the net force acting. In this situation the gravitational force and the parallel component of the weight force are acting in the same direction, so they should be added. a = (1.36m + 4.9m)/m = 6.26 m/s/s (see, this is where the m cancels out!)

Use v = u + at, so 0 = 3.3 - 6.26t ---> t = 0.53 s.

We now need to know how far up the ramp it went:

d = ut + 1/2at^2 = 3.3*0.53 -1/2*6.26*0.53^2 = 1.75 - 0.88 = 0.87 m

The force accelerating it back down the slope will be the parallel component *minus* the frictional force, since the friction always opposes the motion. F = 4.9m - 1.36m = 3.54m N

a = F/m = 3.54m/m = 3.54 m/s/s

Using the same equation as above but with a known distance and the new acceleration:

d = ut + 1/2at^2

0.87 = 0(t) +1/2*3.54*t^2 (since the initial velocity for this section is 0)

this gives t^2 = 0.87/1.77 = 0.49, which gives t = 0.7 s (side note: it makes sense that this is longer than the trip up)

That gives the total time from start to finish as 0.53 + 0.7 = 1.23 s