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

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Homework Help Overview

The problem involves a work-energy scenario where Sam, with a mass of 75 kg, descends a frictionless slope while experiencing a headwind force. The objective is to determine his speed at the bottom of the slope, which is 50 meters high and inclined at 20 degrees.

Discussion Character

  • Exploratory, Assumption checking

Approaches and Questions Raised

  • Participants discuss the application of work and energy principles, questioning the initial setup of the energy equation and the impact of the headwind on the calculations.

Discussion Status

Some participants have provided guidance on reconsidering the terms in the energy equation, particularly regarding the direction of the headwind's force. The original poster has attempted to re-solve the problem and has indicated a change in their approach based on feedback received.

Contextual Notes

There is a mention of the original poster's initial assumption regarding the headwind's effect on the energy calculations, which was not accounted for in their first attempt. The discussion reflects on the importance of correctly interpreting forces in the context of energy work.

Naomi
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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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