Work-Energy Problem: Solve Horizontal Distance Skier Travels

[billu77]

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

 

[tiny-tim]

Welcome to PF!

Hi billu77! Welcome to PF!

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

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! ​

 

[billu77]

Hi billu77! Welcome to PF!

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

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! ​

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

 

[tiny-tim]

Hi billu77!

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

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)

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

do a gh thingy and a µ thingy separately …

what do you get?

 

[billu77]

Hi billu77!

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


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

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

do a gh thingy and a µ thingy separately …

what do you get?

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

 

[tiny-tim]

(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? ​

 

[billu77]

(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? ​

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