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: μ_{s} = 0.6 and μ_{k} = 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 μ_{s}> tan
θ= 3/4. Since μ_{s}= 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 xdirection is aligned up the ramp
slope. (Note that the ramp is a 345 triangle, so sin θ = 0.6 and
cos θ = 0.8).
Steps 24: See transparency.
Step 5: Apply Newton's 2nd Law
ydirection: N = mg cos θ = 0.8 mg
xdirection: ma_{up} = mg sin θ  μ_{k} N =  (0.6+0.4) mg = mg. a_{up} = g
Δx  = 

=  0.80 m 
(a) Distance traveled. Use: v^{2} = v_{o}^{2} + 2 a_{up} Δx
t_{up}  = 

=  0.40 s 
(b) Stopping time. Use:
v = v_{o} + a_{up}t_{up}
xdirection: mg sin θ  μ_{k} N = ma_{down} a_{down} = (0.60.4) g = 0.2 g = 2.0 m/s^{2}
(c) Sliding time down: Use x = ½ a_{down}t^{2}_{down} or 0.8 = t_{down}^{2}. t_{down}= 0.89 s.
(d) Final speed: Use
v^{2}_{final} = v_{o}^{2} + 2
a_{down} Δx = 0 + 2*2*0.8.
v_{final}= 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.