# Speed to give box up incline to reach top (work-energy theorem)

I have come across a problem that I thought I was doing correctly but it seems I am not as my answer seems way too easy and is not the right one either. I don’t need the question worked out for me. I just need someone to give me a nudge in the right direction.

**1. The problem statement, all variables and given/known data**

“You must project a box up an incline of constant slope angle α so that it reaches a vertical distance h above the bottom of the incline. The incline is slippery but there is some friction present.

Use the work-energy theorem to calculate the minimum speed you must give the box at the bottom of the incline so that it will reach the top. Express your answer in terms of g, h, µ, and α.”

**3. The attempt at a solution**

My attempt was using:

(1/2)mv^{2} = mgh + µmg cos(α) (h /sin(α))

with h / sin(α) being the hypotenuse(i.e. the distance up the ramp)

When I try to solve this for v I get to:

v = (2gh + 2µgh cot(α))^{1/2}

However, the answer is supposed to be:

v = (2gh[1 + μ/tan(α)])^{1/2}

Any help would be greatly appreciated. 🙂

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