Skiing up a Hill

  • Thread starter Hypnos_16
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  • #1
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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
 

Answers and Replies

  • #2
tiny-tim
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Homework Helper
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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:)
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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  • #3
153
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so what you're saying is i already found the answer?
 
  • #4
tiny-tim
Science Advisor
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he he! :biggrin:

life sometimes is that simple! :smile:

(oh, they might want you to say whether the work done is positive or negative)
 
  • #5
153
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But wait, wouldn't i need to use the sin 26.8 degrees for the force of gravity since it's down a hill?!
 
  • #6
tiny-tim
Science Advisor
Homework Helper
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Hi Hypnos_16! :smile:
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:
 
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