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: μ_s = 0.6 and μ_k = 0.5. The box is then given an initial velocity up the ramp of 4.0 m/s.

The box slows down as it slides up the ramp and eventually comes to a stop. Will the box remain at rest or will the stop just be for an instant before it slides down the ramp again?

1. The box remains at rest (15/27) (56%)
2. The box slides down the ramp (12/27) (44%)

For the box to remain at rest we need a coefficient of static friction of at least tan(θ), which for this ramp is 3/4. The coefficient of static friction is actually 0.6, not enough to keep the box at rest. So, it slides back down again.

If there is no friction between the box and the ramp the sliding-up part of the trip takes the same time as the sliding-down part of the trip. Here we have friction - how do the two times compare?

1. Sliding up takes more time than sliding down (16/35) (46%)
2. Sliding up takes less time than sliding down (10/35) (29%)
3. It's still the same time even with friction (9/35) (26%)

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) + μ_k 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 = v_o² + 2 a₁ Δx

Defining up as positive gives:

Δx

=

v2 - vo2 2a1

=

0 - 16 -20

=

0.80 m

(b) How long did that take?

v = v_o + a₁t₁

t1

=

v - vo a1

=

0 - 4 -10

=

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) - μ_k 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 = v_ot₂ + 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 = v_o² + 2 a₂ Δx = 0 + 2*2*0.8

This gives a speed at the bottom of 1.8 m/s.