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A skier going up a hill

  • Thread starter omc1
  • Start date
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????


Science Advisor
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 isnt 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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