crate on a plane?

Question

A crate is given an initial speed if 3.0m/s up a 25 degrees plane. How far up the plane will it go? How much time elapses before it returns to its starting point? assume mu=0.17.

Answer 1

v = 3.0m/s, θ = 25, μ = 0.17
let d be the distance it travels up the plane and
let h be the vertical height
so that tanθ = h/d gives h = dtanθ..........(1)
then by conservation of energy
½mv² = mgh + μNd
substituting (1) and noting the normal N = mgcosθ gives
= mg(dtanθ) + μ(mgcosθ)d
= mgd(tanθ + μcosθ)
so
v² = 2gd(tanθ + μcosθ)
gives
d = v² /2g(tanθ + μcosθ) = calculator

,.,.,.,.,.,.,.,.,.

Answer 2

The crate's motion depends on the forces acting on it, so make a list of those forces. Since the motion is along the incline, consider the components of force in the directions parellel and perpendicular to the incline.

There are 3 forces:

1. The force due to gravity (the crate's weight). This equals "mg" and acts downard. The component parallel to the slope is (mg)sin25 (down the slope); and the perpendicular component is (mg)cos25 (pushing into the slope).

2. The normal force, Fn, of the inclined plane pushing perpendicularly against the crate.

3. The force of friction, pushing parallel to the slope and opposite of the direction of motion. Its magnitude is Fn(μ) = Fn(0.17)

The two perpendicular forces ("Fn" and "(mg)cos25") cancel each other out. (We know this because there is no acceleration in the perpendicular direction, hence the net perpendicular force is zero.) If they cancel out, it must be that they have the same magnitude (but opposite directions). Therefore, it must be that:

Fn = (mg)cos25

And this tells us the magnitude of the force of friction:

Friction = Fn(μ) = (mg)cos25(μ)

So now we can look at the net forces in the parallel direction. For this, let's use the convention that the "downslope" direction is positive, and the "upslope" direction is negative.

On the trip UP the slope, the gravity component and the friction component act together (they both point down the slope). So for Leg 1 (upward half) of the trip:

F_net_1 = (mg)sin25 + (mg)cos25(μ)

and the acceleration is:
a1 = F_net_1 / m
a1 = (g)sin25 + (g)cos25(μ)
a1 = (g)(sin25 + μcos25)

The distance it travels upslope is given by the kinematics equation:

d = (v_initial)² / (2(a1))
So:
d = (3.0m/s)² / (2g(sin25+μcos25))

[Edit: this "d" is different from Black Wolf's answer because Black Wolf made a mistake. He said "h/d = tanθ"; but really, h/d = cosθ.]

And the time it takes to go upslope, is given by another kinematics equation:

t1 = v_initial / a1
So:
t1 = (3.0ms) / (g(sin25+μcos25))

Now consider the downslope trip. The net force is _different_ for this half of the trip, because this time the force due to gravity and the force due to friction are acting in _opposite_ directions (the gravity pulls downward; the friction pulls upward). So:

F_net_2 = (mg)sin25 − (mg)cos25(μ)
[notice minus sign because the forces oppose each other]

and the acceleration is:
a2 = F_net_2 / m
a2 = (g)sin25 − (g)cos25(μ)
a2 = (g)(sin25 − μcos25)

To get the time for the downslope trip, use the acceleration (a2) and the distance (d) calculated above, and use this kinematics equation:

t2 = sqrt(2d / a2)

Finally, the time for the total trip is just the sum of t1 and t2:

t_total = t2 + t2