### Block on a Ramp

A block starts from the bottom of a ramp of length 4 m and height 3 m with an initial velocity up the ramp of 4 m/s. The coefficients of friction for the block on the ramp are: μs = 0.6 and μk = 0.5.

The block slows as it slides up the ramp and eventually stops. Will the block remain at rest or will it slide down the ramp again?  (Hint:  Find tan θ.)

1. The block remains at rest
2. The block slides down the ramp

The block remains at rest if μs> tan θ= 3/4. Since μs= 0.6, the block backslides.

Now compare the times for sliding up and sliding down:

1. Sliding up takes more time than sliding down.
2. Sliding up takes less time than sliding down.
3. Both times equal, even with friction.

To find the times use DID TASC:
1. Diagram and coordinate systems
2. Isolate the systems
3. Draw all forces acting
4. Take components
5. Apply F=ma and constraints
6. Solve
7. Check!

Step 1: Diagram and coordinate system

Use a coordinate system in which the x-direction is aligned up the ramp slope. (Note that the ramp is a 3-4-5 triangle, so sin θ = 0.6 and cos θ = 0.8).

Steps 2-4: See transparency.

Step 5: Apply Newton's 2nd Law

Sliding Up:

y-direction:     N = mg cos θ = 0.8 mg

x-direction:     maup = -mg sin θ - μk N = - (0.6+0.4) mg = -mg.     aup = -g
Δx =
 v2 - vo2 2aup
= 0.80 m

(a) Distance traveled. Use:  v2 = vo2 + 2 aup Δx
tup =
 v - vo aup
= 0.40 s

(b) Stopping time. Use:  v = vo + auptup

Sliding Down:

x-direction:     mg sin θ - μk N = madown   adown = (0.6-0.4) g = 0.2 g = 2.0 m/s2

(c) Sliding time down: Use x = ½ adownt2down    or    0.8 = tdown2.       tdown= 0.89 s.

(d) Final speed: Use v2final = vo2 + 2 adown Δx = 0 + 2*2*0.8.      vfinal= 1.8 m/s.

Intuitive explanation:

Going up: friction and gravity work together; the block decelerates quickly.
Going down: friction and gravity work in the opposite direction; the block accelerates slowly.

Sliding up takes less time.