How to Calculate Distance Traveled on a Slope with Kinetic Friction

[hoseA]

An Olympic skier moving at 28 m/s down
a 24 degrees slope encounters a region of snowless
ground of coefficient of kinetic friction 0.67.
The acceleration of gravity is 9.8 m/s^2
How far down the slope does she travel
before coming to a halt? Answer in units of
m.

I came up with an equation of:

mgsin(theta) - Fk*d = ma

I would solve for "d" but I'm not sure how to get the "a".

help appreciated.

TIA.

 

[civil_dude]

Try this,

The sum of the forces parallel to the slope is mgcos62(gravity) - mgUk(friction)

mg(Uk-cos62) = - 0.21 mg = -2.06m/s^2 * m

so your acceleration, a, is 2.06 m/s^2 going against the skier.

From there you use t = dv/a = 28/2.06 so t = 13.6 s

then L traveled is L = 1/2 a t^2 = 44.8 m

Or something like that.

 

[hoseA]

Try this,
The sum of the forces parallel to the slope is mgcos62(gravity) - mgUk(friction)
mg(Uk-cos62) = - 0.21 mg = -2.06m/s^2 * m
so your acceleration, a, is 2.06 m/s^2 going against the skier.
From there you use t = dv/a = 28/2.06 so t = 13.6 s
then L traveled is L = 1/2 a t^2 = 44.8 m
Or something like that.

Thanks a lot. Your math is off in the last step but you had the right concepts.

I did mgsin(theta)-Ukmgcos(theta) = ma

a = 2.01...

t = 28/2.01...

L = .5at^2 = 195.25 (not precise but within 1% margin of error)

Thanks again. :)