| v2 - vo2 |
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| 2a1 |
| 0 - 16 |
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| -20 |
A box is placed at the bottom of a ramp that measures 3.00 m vertically by 4.00 m horizontally. The coefficients of friction for the box and the ramp are: μs = 0.6 and μk = 0.5. The box is then given an initial velocity up the ramp of 4.0 m/s.
One way to think about this is in terms of average velocity. On the way up the average velocity is 2.0 m/s up (the average of the initial velocity, 4.0 m/s up, and the final velocity, 0). On the way down the average velocity is the average of the initial velocity, 0, and the final velocity, which must be less than 4.0 m/s down because of friction. So, the average velocity on the way down is less than 2.0 m/s down.
Covering the same distance with a larger average velocity takes less time, so the sliding-up part of the trip takes less time.
Let's work some things out, including:
(a) How far does the box slide up the ramp?
(b) How long does it take to slide up, and to slide down?
(c) What is the box's speed when it gets back to the bottom of the ramp again?
Let's use g = 10.0 m/s2.
First, note that the ramp is a 3-4-5 triangle, so sin(θ) = 0.6 and cos(θ) = 0.8.
Find the acceleration on the way up and on the way down. Sketch a free-body diagram.
Use a coordinate system aligned with the ramp. In the y-direction, perpendicular to the ramp, we get:
N = mg cos(θ) = 0.8 mg
For the sliding up part of the trip we have an acceleration down the ramp given by:
mg sin(θ) + μk N = ma1
ma1 = 0.6 mg + 0.5 * 0.8 mg
a1 = 0.6 g + 0.4 g = g = 10 m/s2, directed down the ramp
(a) To figure out how far it goes we can use:
v2 = vo2 + 2 a1 Δx
Defining up as positive gives:
| Δx | = |
|
= |
|
= | 0.80 m |
(b) How long did that take?
v = vo + a1t1
| t1 | = |
|
= |
|
= | 0.40 s |
What about going down? We need to find the acceleration for the down part of the trip. Applying Newton's second law to the sliding-down part of the trip gives:
mg sin(θ) - μk N = ma2
a2 = 0.6 g - 0.4 g = 0.2 g = 2.0 m/s2, directed down the ramp
To figure out how long it takes to slide down, use:
x = vot2 + 0.5 a2t22
The initial velocity is zero on the way down, so (with down positive):
0.8 = 0 + t22
t2= 0.89 s.
(c) To find out the speed of the box at the bottom of the ramp, we can use:
v2 = vo2 + 2 a2 Δx = 0 + 2*2*0.8
This gives a speed at the bottom of 1.8 m/s.