Law of Conservation of Energy of a downhill skier

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Homework Help Overview

The discussion revolves around the application of the law of conservation of energy in a scenario involving a skier moving down a slope and then up another hill. The problem involves calculating the skier's speed at the top of the second hill, given specific distances and angles of inclination.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to apply the conservation of energy principle but expresses confusion regarding the interpretation of distances as heights. Some participants suggest using trigonometric functions to find the actual heights corresponding to the given distances and angles.

Discussion Status

Participants are actively exploring different interpretations of the problem, particularly regarding the relationship between distance traveled and height. Some guidance has been offered on using trigonometric functions to resolve the confusion about height calculations.

Contextual Notes

There is an ongoing discussion about the implications of the given distances and angles, with participants questioning the assumptions made about the problem setup. The original poster acknowledges a known answer but struggles to reconcile it with their calculations.

darkmasterz8
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1. A skier is pushed from the top of a hill so that he starts moving down the hillside sloped at 27.6° to the horizontal with an initial speed of 0.434 m/s. After traveling 80.4 m, he reaches the bottom of the valley. Due to inertia, he then continues 70.4 m up another hillside sloped at 20.7° to the horizontal. What is the skier's speed when he reaches the top of the hill? Assume that you can neglect friction.
2. KE(b)+PE(b)=KE(a)+PE(a)
KE(b)+PE(b)=KE(a)+PE
1/2*m*(.434^2)+m*9.8*80.4=1/2*m*v^2+m*9.8*70.4---X out all m since they're the same both sides

.09+787.9=1/2*v^2+689.9
788=1/2*v^2+689.9
98.11=1/2*v^2
196=v^2
14=v

I already know the answer is 15.6 but I don't seem to get it through the work I've done. What I'm confused on is if 80.4 and 70.4 are the height? The problem makes it sound more like distance which doesn't make sense to me. I'm pretty sure that is the formula for law of conservation of energy.
 
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yes 80.4 and 70.4 are the distances that the person travels...not the height, you will have to use the angle of inclination to find the actual height

and yes, you will encounter a lot of problems that give you distances to make your life harder
 
Well to use angle of inclination to find the height for initial, you do cosine 27.6=x/80.4 where x = 71.2 meters. The same thing with the after part. But when I plug in the height into the above equation, I still don't get it.

I'm not sure which of the three to use; cosine/tangent/sine because I don't know where the angle belongs. It says 27.6 to the horizontal so I assume the angle is at the top and you use cosine to find the adjacent given the hypotnuse of 80.4.

EDIT:
OOO I think I just got it. It's sine to get it.

Sine 27.6=x/80.4 where x=37.25
Sine 20.7=x/70.4 where x=24.77

Sub it in the above equation and I get 15.6 m/s.

Thanks for telling me about height/distance :)
 
Last edited:
draw a triangle with a right angle, the distance is the hypotenuse and the angle of inclination is opposite the height

this might be a good way to remember it ;)
 

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