How Does Kinetic Friction Affect a Skier's Ascent Up a Hill?

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Homework Help Overview

The discussion revolves around a physics problem involving a skier coasting up a hill, focusing on the effects of kinetic friction on the skier's ascent. The problem includes parameters such as the skier's mass, initial and final speeds, the angle of the hill, and the distance traveled up the slope.

Discussion Character

  • Exploratory, Mathematical reasoning, Assumption checking

Approaches and Questions Raised

  • Participants discuss the relationship between kinetic energy, potential energy, and the work done by kinetic friction. There are attempts to derive expressions for energy changes and to calculate the work done by friction based on energy conservation principles.

Discussion Status

Some participants have provided calculations and expressed confusion about discrepancies in their results. There is an acknowledgment of the need to clarify the definitions and relationships between the energies involved, with some guidance offered regarding the nature of the problem.

Contextual Notes

Participants are working under the constraints of a homework assignment, which may limit the information they can use or the methods they can apply. There is a focus on understanding the work done by kinetic friction rather than calculating the force or coefficient of friction directly.

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Homework Statement

A 61.2kg skier coasts up a snow-covered hill that makes an angle of 25.2o with the horizontal. The initial speed of the skier is 8.34m/s. After coasting a distance of 1.97m up the slope, the speed of the skier is 3.43m/s. Calculate the work done by the kinetic frictional force that acts on the skis

Homework Equations

w= ΔE
ke= 1/2mv^2 pe=mgh

The Attempt at a Solution

h=dsin(theta) so i found mu(k) = -1/2mv^2 (final)+1/2mv^2(initial)-mgdsin(theta) all divided by Nd where N=mgcos(theata)

i got 0.654 J...not working?
 
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Okay, going a distance d up the slope, at angle \theta, the skier has gone vertically dsin(theta) meters and his potential energy has increased by mgd sin(theta) Joules. The skier's original kinetic energy was (1/2)mv^2(initial) so that increase in potential energy causes a decrease in kinetic energy to (1/2)mv^2(initial)- mgd sin(theta). Since the final kinetic energy is, in fact, (1/2)mv^2(final), they must be an additional loss of energy of (1/2)mv^2(initial)- mgdsin(theta)- (1/2)mv^2(final). That is the "work done by the kinetic frictional force"

The problem does NOT ask for the force itself or the coefficient of friction, which is what you are calculating.
 
i got 699.31 J which isn't right, what am i missing i used (1/2)mv^2(initial)- mgdsin(theta)- (1/2)mv^2(final)...
 
i found my problem thanks
 

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