**Delivering Rescue Supplies**

Part A = v = sqrt(2gh(μ_{k}cot(α) + 1))

**Solution Below:**

You are a member of an alpine rescue team and must project a box of supplies, with mass m, up an incline of constant slope angle α so that it reaches a stranded skier who is a vertical distance h above the bottom of the incline. The incline is slippery, but there is some friction present, with kinetic friction coefficient μ

_{k}.

Part A

Use the work-energy theorem to calculate the minimum speed v that you must give the box at the bottom of the incline so that it will reach the skier.

Express your answer in terms of some or all of the variables m, g, h, μ

Express your answer in terms of some or all of the variables m, g, h, μ

_{k}, and α.There are several ways to approach this problem. The easiest is just to figure out the forces from gravity and kinetic friction, and make sure the velocity is high enough to overcome them. When you solve, you actually remove mass from the equation. The answer that mastering physics is looking for is:

v = sqrt(2gh(μ_{k}cot(α) + 1))

I am still confused on how to solve this problem

In a nutshell, you need to equate kinetic energy at the bottom of the slope with the energy required to reach the top. The energy required will be equal to the gravitational energy, plus the amount of work needed to overcome friction as the box moves up the hill.

1/2mv^2 = mgh + (μmgcos(α) * L) (where L is the length of the incline)

Mass cancels, so you end up with:

1/2v^2 = gh + (μgcos(α) * L)

This is equivalent to: (note that this step uses a trig identity to convert (cos(α) * L) to (cot(α) * h))

1/2v^2 = gh + (μgcot(α) * h)

1/2v^2 = gh + (gh * μcot(α))

Factor out gh:

1/2v^2 = gh * (1 + μcot(α))

Now isolate v^2, and then v:

v^2 = 2gh * (1 + μcot(α))

v = sqrt(2gh * (1 + μcot(α)))