Work-Energy Problem: Solve Horizontal Distance Skier Travels

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A skier starts from rest on a frictionless incline of 20.0 m height and 20 degrees angle, encountering friction on both the incline and horizontal surface. The discussion focuses on calculating the horizontal distance traveled before the skier comes to rest, emphasizing the need to treat gravitational and frictional work separately. Initial confusion arose regarding the use of sine and cosine in calculations, with clarification provided on the correct application of forces. After applying the correct equations, the user successfully determined the horizontal distance using the coefficient of friction and gravitational force. The conversation highlights the importance of understanding the physics principles involved in work-energy problems.
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Homework Statement





Problem A skier starts from rest at the top of a frictionless incline of height 20.0 m which makes an angle of 20 degrees with the horizontal. As the bottom of the incline, the skier encounters a horizontal surface where the coefficient of kinetic friction between skis and snow is 0.210.

Find the horizontal distance the skier travels before coming to rest if the incline also has a coefficient of kinetic friction equal to 0.210.

The Attempt at a Solution



-mg(mu) cos 20 = -1/2 mv ^2
-g cos 20 = -1/2 v ^2

unable to get the right answer...pls help
 
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Welcome to PF!

Hi billu77! Welcome to PF! :smile:

(have a mu: µ and try using the X2 tag just above the Reply box :wink:)

I htink you're getting rather confused …

i] the height will be sin, not cos

ii] the friction is only on the horizontal

Try again! :smile:
 


tiny-tim said:
Hi billu77! Welcome to PF! :smile:

(have a mu: µ and try using the X2 tag just above the Reply box :wink:)

I htink you're getting rather confused …

i] the height will be sin, not cos

ii] the friction is only on the horizontal

Try again! :smile:


thanks...i tried again with sin instead of cos and it didnt work out...there is friction on both surfaces...as it says on the last line "if the incline has a co-efficient of kinetic friction"...can u solve it showing all the steps...thanks
 
Hi billu77! :smile:

(we don't give out answers here at PF :wink:)
billu77 said:
thanks...i tried again with sin instead of cos and it didnt work out...there is friction on both surfaces...as it says on the last line "if the incline has a co-efficient of kinetic friction"...can u solve it showing all the steps...thanks

ah, I thought that was just part b) … didn't realize there wasn't a part a) :blushing:

ok … you can't combine the gravitational work done and the friction work done into one big µgh thingy :-p

do a gh thingy and a µ thingy separately …

what do you get? :smile:
 
tiny-tim said:
Hi billu77! :smile:

(we don't give out answers here at PF :wink:)


ah, I thought that was just part b) … didn't realize there wasn't a part a) :blushing:

ok … you can't combine the gravitational work done and the friction work done into one big µgh thingy :-p

do a gh thingy and a µ thingy separately …

what do you get? :smile:

I have no clue about the method that u mentioned...atleast if you could show me the equations, i can try to work out the solution...gh alone would be 9.8 x 20= 196...i am attaching an image of the problem...would appreciate if you could provide the equations and i can take it from there...thanks
 

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(I can't see your picture yet, but anyway …)

The gravitational work done is mgh, because mg downward is the force, and if we "dot" it with the displacement, that means multiplying it by the difference in height, h.

The frictional work done must be calculated separately, and is the friction force "dot" the displacement …

what is that? :smile:
 
tiny-tim said:
(I can't see your picture yet, but anyway …)

The gravitational work done is mgh, because mg downward is the force, and if we "dot" it with the displacement, that means multiplying it by the difference in height, h.

The frictional work done must be calculated separately, and is the friction force "dot" the displacement …

what is that? :smile:


thanks a lot...i got the correct answer using the equation:

0.210 x 9.8 x height = 0.210 cos 20 x horizontal distance of incline + 0.210 x 9.8 x distance traveled totally
 
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