for all you physicists a?

[bah]

an olympic skier moving at 20.0 m/s down a 30.0* slope encounters a region of wet snow and slides 145 m before coming to a halt. what is the coefficient of friction between the skis and the snow?

every equation i have needs a weight (more accurately the normal force) and from all i can tell i cant find the normal force, something obviously has to cancel out somewhere, but i cant find what, this is due tomorrow so help how would be nice ( i know a majority of you drive comeros, and this is more of a firebird problem, but do your best, haha)

[BLLDOGG]

Re: for all you physicists a?

yeah, it is a problem when all you firebird kids need help with your homework.

[hockeyman]

Re: for all you physicists a?

a FIREBIRD moving at 20.0 m/s down a 30.0* slope encounters a region of wet snow and slides 145 m before coming to a halt.

= A wall

hahaha

camaro's rule!:banana:

[3.4 slow to go]

Re: for all you physicists a?

I take it you're using the demouveres theorem for this, correct? or am I jsut having flashbacks of my fluid mechanics class??????

[bah]

Re: for all you physicists a?

this is high school physics, and from the wording i am assuming i am finding kinetic friction before the rest, i have found the acceleration is -1.592691457 m/s^2 the time is 12.557365 seconds the x componet is 125.5736 m and the y is 72.5 m but all the equations i have i need the normal force to find the friction, and the normal force is pretty difficult to find without a mass coefficient of kinetic friction = force of friction/normal force i cannot find either force without knowing the mass because force = mass x acceleration

[3.4 slow to go]

Re: for all you physicists a?

i haven't had to mess w/that shyt for over 13yrs, thanks

[notsonic]

Re: for all you physicists a?

it took 14.49seconds.

acceleration due to friction is -1.38.

and um yeah.

i cant believe i cant figure this out.

[airattack111]

Re: for all you physicists a?

= A wall

hahaha

camaro's rule!:banana:

+1
.

[notsonic]

Re: for all you physicists a?

i got .16

[southalabama97]

Re: for all you physicists a?

Is the 30* to the horizontal or vertical? I'll see if I can come up with something.

[bah]

Re: for all you physicists a?

its to the horizontal, and i was able to solve it for a constant velocity, which was like .56 or something, and using deductive reasoning because it had a negative velocity that means that friction overcame gravity, which means it must be a greater coefficient of friction to stop the skier, it is multiple choice a. 0.740 b. 0.540 c. 0.116 and d. 0.470 and if it must be greater than .56 that would just leave a, but i might be completely wrong, and i dont want bs answers so if you dont know what you are doing i would prefer you kept your mouth shut, those of you that do know what you are doing i appreciate this a lot, i have just been thinking about this problem for like 2 hours

[notsonic]

Re: for all you physicists a?

i think itd be easier if we knew what chapter you were up to. :)

but this is how i did it...which i dont know if its really correct.

work done by friction = change in kinetic energy of the skiier.

so

w=.5mv^2

F(friction)*145=200m

un*145=200m

u(mgcos30)145=200m

mass drops out

u(9.8 cos30)145=200

u1225.25=200

u=.16

BUT since its not a choice i dunno what i mightve f'ed up. im probably missing a cos or something.

[southalabama97]

Re: for all you physicists a?

I did it a different way and also came up with .16. I took the sum of forces at the skier in the x direction and y direction and solved. I put the axis on the 30*plane so that it wouldn't matter the angle.

Used the forumula

Vf^2 = Vi^2 +2a(xf-xi)
since Vf and xi are 0 it reduces to

0 = Vi^2 +2axf

Since a is -ug

0 = Vi^2 +2ugxf

u = Vi^2/2gxf = 20^2/2*9.81*145

u = 0.162