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

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SUMMARY

In the work-energy problem involving Sam, who has a mass of 75 kg and descends a 50 m high frictionless slope at a 20-degree angle, the correct approach to finding his speed at the bottom involves accounting for the headwind exerting a horizontal force of 200 N. Initially, the incorrect application of the work-energy principle led to an erroneous velocity calculation of 41 m/s. Upon reevaluation, the correct formula incorporates the work done by the headwind as negative, resulting in a final speed of approximately 16 m/s.

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  • Familiarity with gravitational potential energy (mgh)
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  • Basic trigonometry for slope calculations
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This discussion is beneficial for physics students, educators, and anyone interested in mastering the application of the work-energy principle in mechanics, particularly in scenarios involving external forces like wind resistance.

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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?
 
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Naomi said:
200N(50/tan(20))=1/2mvf^2 -mghi
have another think about that. Which term supplied energy, and which terms absorbed it?
 
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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!
 

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