Find distance before rest given length and angle of incline

AI Thread Summary
A skier starts from rest on a 9.1° incline, with a hill length of 250 m and a friction coefficient of 0.0750. The height of the hill is calculated to be 39.5 m, leading to a velocity of 27.8 m/s at the bottom. Discussions highlight the need to account for friction on both the slope and the horizontal portion when calculating the distance the skier glides before stopping. The work-energy theorem is suggested as a method to find the stopping distance, but the mass of the skier complicates the calculations. Participants emphasize the importance of correctly applying energy principles and friction to arrive at an accurate solution.
mandy9008
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Homework Statement


A skier starts from rest at the top of a hill that is inclined at 9.1° with the horizontal. The hillside is 250 m long, and the coefficient of friction between snow and skis is 0.0750. At the bottom of the hill, the snow is level and the coefficient of friction is unchanged. How far does the skier glide along the horizontal portion of the snow before coming to rest?


The Attempt at a Solution


250sin9.1= 39.5 (height of hill)

v2 = 2gh
v2 = 2 (9.8 m/s2)(39.5m)
v= 27.8 m/s

d= v2 / 2 u g
d= 27.8 m/s / 2 (0.0750) (9.8 m/s2)
d= 18.9 m
 
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Do you know the mass of the skier?
 
No, i do not know the mass of the skier.
 
That is weird because friction force depends on the normal force which depends on the mass of the object.
By the way, you did not take into account friction here:

v2 = 2gh
v2 = 2 (9.8 m/s2)(39.5m)
v= 27.8 m/s
 

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I used the coefficient of kinetic friction when I solved for the distance
 
Yes but there is friction in the slope also. In my opinion this problem should be split into two parts, first find the velocity and the end of the slope then find the distance required to stop. The later one can be solved using work energy theorem, but I still don't see how to do this without the mass. Sorry, maybe someone else can help you more. Good luck.
 
all I can give you is theory atm.. .

the skier gains kinetic energy on the way down. at the bottom of the slope Ek = Eg-W

W= (coefficient of kinetic friction * mass * gravity)(distance along the slope)

so then you just take Ek at the bottom and sub in for W=uFn*d

W/uFn = d
 
how do I get the W?
 
oh just read the question

mgh - umg = 1/2mv^2

9.8*h-0.075*9.8=1/2v^2

9.725(39.53)=1/2v^2

384.52*2=v^2

27.73m/s
 
  • #10
W/uFn = d

27.73/0.075(mg)=d

27.73 / (0.735m) = d

i have no clue how to get beyond this point.

lol try calculating mass
 
  • #11
joshmdmd said:
W/uFn = d

27.73/0.075(mg)=d

27.73 / (0.735m) = d

i have no clue how to get beyond this point.

lol try calculating mass

this is wrong.. stupid mistake

27.73 has to be converted to energy .. gl
 
  • #12
okay so I was just playing around with these numbers and I divided 27.73m/s / 9.8m/s^2 and got 2.83m. When I entered this number, the response told me that I was off by a multiple of 10. I only have one submission left before I reach the max, so I want to get this one right. Any ideas?
 
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