Skiing Up a Hill: Work Done by Friction

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Homework Help Overview

The problem involves a skier coasting up a hill, analyzing the work done by kinetic friction. The skier's mass, initial and final speeds, and the distance traveled up the slope are provided, along with the angle of the hill.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • The original poster attempts to calculate the total work done using the work-energy principle but struggles to isolate the work done specifically by friction. Some participants suggest that the only forces acting are gravity and friction, with gravity already accounted for in the energy calculations.

Discussion Status

The discussion is ongoing, with participants exploring the implications of the calculations and questioning the treatment of gravitational force in the context of the work-energy equation. There is no explicit consensus on the interpretation of the results yet.

Contextual Notes

Participants note the need to clarify whether the work done by friction is positive or negative, and there is a discussion about the correct application of gravitational force components in the calculations.

Hypnos_16
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Homework Statement



A 69.7-kg skier coasts up a snow-covered hill that makes an angle of 26.8° with the horizontal. The initial speed of the skier is 6.16 m/s. After coasting 1.98 m up the slope, the skier has a speed of 3.95 m/s. Calculate the work done by the kinetic frictional force that acts on the skis.

m = 69.7
Theta = 26.8 degrees
v1 = 6.16 m/s
d = 1.98m
v2 = 3.95 m/s


Homework Equations



I've found the total work done by using Wnc = Ef - Ei
but that gives me the wok of all the forces and i don't know how to pinpoint just the force done by friction


The Attempt at a Solution


∑W = Ef - Ei
∑W = (1/2mv^2) + (mgh) - (1/2mv^2) + (mgh)
∑W = (34.9)(3.95^2) + (69.7)(9.81)(1.98) - (34.9)(6.16^2) + (69.7)(9.81)(0)
∑W = (544) + (1353.8) - (1324) + (0)
∑W = 1897.8 - 1324
∑W = 573.8 J
 
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Hi Hypnos_16! :smile:

(have a theta: θ and try using the X2 and X2 icons just above the Reply box :wink:)
Hypnos_16 said:
I've found the total work done by using Wnc = Ef - Ei
but that gives me the wok of all the forces and i don't know how to pinpoint just the force done by friction

The only forces (apart from gravity, which you've included in your E) are the normal force, N, which does no work, and the https://www.physicsforums.com/library.php?do=view_item&itemid=39"

so all the https://www.physicsforums.com/library.php?do=view_item&itemid=75" (in this case) is by friction! :smile:
 
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so what you're saying is i already found the answer?
 
he he! :biggrin:

life sometimes is that simple! :smile:

(oh, they might want you to say whether the work done is positive or negative)
 
But wait, wouldn't i need to use the sin 26.8 degrees for the force of gravity since it's down a hill?!
 
Hi Hypnos_16! :smile:
Hypnos_16 said:
But wait, wouldn't i need to use the sin 26.8 degrees for the force of gravity since it's down a hill?!

You already used the work done by gravity when you calculated the energy …

you included mgh in your E …

in coordinates, that's (x,y,h).(0,0,g), ie minus distance "dot" force …

https://www.physicsforums.com/library.php?do=view_item&itemid=269" by that force …

in the work-energy equation, you can either include gravity in the PE (on one side), or in the work done (on the other side), but not both! :wink:
 
Last edited by a moderator:

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