Question

To practice Problem-Solving Strategy 10.1 for energy conservation problems. A sled is being held at rest on a slope that makes an angle θ with the horizontal. After the sled is released, it slides a distance d1 down the slope and then covers the distance d2 along the horizontal terrain before stopping. Find the coefficient of kinetic friction μk between the sled and the ground, assuming that it is constant throughout the trip.

Answer 1

Answer:

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]

Explanation:

To solve this, let's use the work/energy theorem which states that: The change in an object's Kinetic energy is equal to the total work (positive and/or negative) done on the system by all forces.

Now, in this question, the change in the object's KE is zero because it starts at rest and ends at rest. (ΔKE = KE_final − KE_initial = 0). Thus, it means the sum of the work, over the whole trip, must also be zero.

Now, if we consider the work done during the downhill slide,there will be three forces acting on the sled:

1. Weight (gravity). This force vector has magnitude "mg" and points points straight down. It makes an angle of "90°–θ" with the direction of motion. Thus;

Wgrav = (mg)(d1)cos(90°–θ)

From trigonometry, we know that cos(90°–θ) = sinθ, thus:

Wgrav = (mg)(d1)sin(θ)

2. Normal force, Fn=(mg)cosθ. This force vector is perpendicular to the direction of motion, so it does zero work.

3. Friction, Ff = (Fn)μk = (mg) (cosθ)μk and it points directly opposite of the direction of motion,

Thus;

Wfric = –(Fn)(d1) = –(mg)(cosθ)(μk)(d1)

(negative sign because the direction of force opposes the direction of motion.)

So, the total work done on the sled during the downhill phase is:

Wdownhill = [(mg)(d1)sin(θ)] – [(mg)(cosθ)(μk)(d1)]

Now, let's consider the work done during the "horizontal sliding" phase. The forces here are:

1. Gravity: it acts perpendicular to the direction of motion, so it does zero work in this phase.

2. Normal force, Fn = mg. It's also perpendicular to the motion, so it also does zero work.

3. Friction, Ff = (Fn)(μk) = (mg)(μk). Thus; Wfric = –(mg)(μk)(d2) (negative because the direction of the friction force opposes the direction of motion).

The total work done during this horizontal phase is:

Whoriz = –(mg)(μk)(d2)

Hence, the total work done on the sled overall is:

W = Wdownhill + Whoriz

= (mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2)

I have deduced that the total work is zero (because change in kinetic energy is zero), thus;

(mg)(d1)sin(θ) – (mg)(cosθ)(μk)(d1) – (mg)(μk)(d2) = 0

Now, let's make μk the subject of the equation:

First of all, divide each term by mg;

(d1)sin(θ) – (cosθ)(μk)(d1) – (μk)(d2) = 0

Rearranging, we have;

(d1)sin(θ) = (cosθ)(μk)(d1) + (μk)(d2)

So,

(d1)sin(θ) = [(cosθ)(d1) + (d2)](μk)

And

μk = (d1)sin(θ) / [(cosθ)(d1) + (d2)]