This problem has got me stumped. I need the answer to make an A on a test

  • Thread starter tdcman92
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  • #1
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1. Homework Statement
this straight from a textbook.
A skier starts from rest at the top of a hill that is inclined at 10.5degrees with the horizontal. The hillside is 200.0 m long, and the coefficient of friction between the snow and the skies is 0.075. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier move along the horizontal portion of thesnow before coming to rest?


2. Homework Equations



3. The Attempt at a Solution
i tried different thing but i cant enderstand how the problem can be solved without the mass of the skier.
 
Last edited:

Answers and Replies

  • #2
ehild
Homework Helper
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How would you solve the problem if you knew the mass of the skier?

ehild
 
  • #3
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How would you solve the problem if you knew the mass of the skier?

ehild

idk how
 
  • #4
161
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try resolving the forces up and down the slope
 
  • #5
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use the conservation of mechanical energy, i think the masses should cancel out
 
  • #6
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i think this is the way to do it:

mgh = 1/2mv^2 + Wf
mgh = 1/2mv^2 + mu(Fn)cos theta (r)
mgh = 1/2mv^2 + mu(mg)cos theta (r) *h can be found by using sin of 10.5
Solve for Vf = ?

plug in Vf as Vi
1/2mv^2 = mg(d) , in this case d is the distance you want...

and i think you should get the answer
 
  • #7
11
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i think this is the way to do it:

mgh = 1/2mv^2 + Wf
mgh = 1/2mv^2 + mu(Fn)cos theta (r)
mgh = 1/2mv^2 + mu(mg)cos theta (r) *h can be found by using sin of 10.5
Solve for Vf = ?

plug in Vf as Vi
1/2mv^2 = mg(d) , in this case d is the distance you want...

and i think you should get the answer

i dont know what mu(Fn)cos theta (r) or mu(mg)cos theta (r) means.
i suck a physics.
 
  • #8
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u can use the law of energy conversion here, what do u think the initial energy that the man has?
 
  • #9
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u can use the law of energy conversion here, what do u think the initial energy that the man has?

i redid it and i figured that the vf is 26.72m/s at the bottom of the hill

now im on the horizontal part of the equation.
the skier starts from 26.72m/s and the coefficient of friction is 0.075
i still need help i feel so dumb
 
Last edited:
  • #10
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the coeffiecient of friction is mu and theta is basically the angle, bottom part of the triangle, and to find use your SohCahToa
 
  • #11
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i redid it and i figured that the vf is 26.72m/s at the bottom of the hill

now im on the horizontal part of the equation.
the skier starts from 26.72m/s and the coefficient of friction is 0.075
i still need help i feel so dumb

can u show us ur calculation? because if u have found the speed at the botton of the hill, then u are supposed to be able to find the rest of the question
 

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