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A skier going up a hill

  • Thread starter omc1
  • Start date
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1. The problem statement, all variables and given/known data 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

2. Relevant equations w= ΔE
ke= 1/2mv^2 pe=mgh

3. 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????
 

HallsofIvy

Science Advisor
41,625
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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.
 
100
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i got 699.31 J which isnt right, what am i missing i used (1/2)mv^2(initial)- mgdsin(theta)- (1/2)mv^2(final)....
 
100
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i found my problem thanks
 

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