Sam, Whose mass is 75kg(Work Energy Problem)

[Naomi]

Sam, whose mass is 75kg, starts down a 50-m-high, 20 degree frictionless slope. A strong headwind exerts a horizontal force of 200N on him as he skies. Use work and energy to find Sam's speed at the bottom.

Hi! I'm new to posting on this website but thought I'd give it a go! I would really appreciate help with this problem. I know the answer is supposed to be around 16m/s, but for some reason I am getting an answer that is too high of a velocity. Here is my attempted solution.

W = ΔKE+ΔU
Given he starts from rest, we know that,

W=KEf+Ui

Fx=200N
W=200N(50/tan(20))

200N(50/tan(20))=1/2mvf^2 -mghi

1712.6=Vf^2
Vf=√(1712.6) =41m/s

However, this solution is not correct. Where did I go wrong?

 

[Drakkith]

Thread has been moved from the General Physics forum.

 

[haruspex]

200N(50/tan(20))=1/2mvf^2 -mghi

have another think about that. Which term supplied energy, and which terms absorbed it?

 

[Naomi]

Okay, I re-solved the problem and got the correct answer. My method seemed to be correct. However, when I initially solved the problem, i did not account for the direction of the headwind (going against Sam). Because of this, I was getting an incorrect answer. Re-solved, my solution looked more like this:
W = KEf-Ui
W= 1/2mVf^2-mghi
W+mghi= 1/2mVf^2
Vf= sqrt((W+mghi)/(.5m))

My formula was correct initially, however I had to solve it with W being (-200N)(50/tan(20)), or -27474 rather than +27474 and all other values remaining the same.Thank for the help! The response definitely prompted me to look at second look at the problem from a different standpoint!

 

[HalfLostAndSearching]

Hi there! Your approach is on the right track, but there are a few mistakes in your calculations. Let me walk you through the correct solution:

First, let's define the initial and final points for Sam's motion. The initial point is at the top of the slope, where Sam has no kinetic energy (KEi = 0) and potential energy (PEi = mgh). The final point is at the bottom of the slope, where Sam has some final velocity (vf) and no potential energy (PEf = 0).

Next, we can use the work-energy theorem to relate the work done by the wind (W) to the change in kinetic energy (ΔKE) and potential energy (ΔPE).

W = ΔKE + ΔPE

Since we know that ΔPE = -mgh (the potential energy decreases as Sam goes down the slope), we can rewrite the equation as:

W = ΔKE - mgh

Now we can plug in the values given in the problem:

W = 200N * 50m/tan(20°) - 75kg * 9.8m/s^2 * 50m

W = 1712.6J

Note that the work done by the wind (W) is equal to the change in kinetic energy (ΔKE) because there is no friction acting on Sam. This means that:

ΔKE = 1712.6J

Finally, we can use the definition of kinetic energy to find the final velocity:

ΔKE = 1/2mvf^2
1712.6J = 1/2 * 75kg * vf^2

vf = √(1712.6J * 2 / 75kg) = 15.9m/s

As you can see, the final velocity is around 16m/s, which is the correct answer. The mistake in your solution was that you used the work done by the wind (W) as the initial kinetic energy (KEi), which is not correct. Instead, we should use the work done by the wind to find the change in kinetic energy (ΔKE). I hope this helps clarify things for you. Keep up the good work!