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.)
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:
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
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
(a) Distance traveled. Use: v2 = vo2 + 2 aup Δx
(b) Stopping time. Use:
v = vo + auptup
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.
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.