# Kinetic friction - finding time (1 Viewer)

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#### inner08

Hi,

I'm not sure if I did this right but here is the word problem and the work i've done. If I made any errors, i'd love to know about them!

A sight seen on many bunny hills across Ontario is young skiers pushing on ski poles and gliding down a slope until they come to rest. Observing from a distance, you note a young person (approximately 25 kg) pushing off with the ski poles to give herself an initial velocity of 3.5 m/s. If the inclination of the hill is 5.0 degrees and the coefficient of kinetic friction for the skis on dry snow is 0.20, calculate

a) the time taken for the skier to come to a stop

For a:

Fa = Fgh + Ff
= Fg sin 5 (degrees) + ukFn

Fn = Fg cos 5 (degrees)
= mg cos 5 (degrees)

Fa = Fg sin 5(degrees) + ukmg cos 5 (degrees)
= mg sin 5(degrees) + ukmg cos 5(degrees)
= (25kg) (9.8m/s^2)(0.08716) + (0.20)(25kg)(9.8m/s^2)(0.9962)
= 70N

F = ma
a = 70N / 25kg
= 2.8 m/s^2

delta t = (Vf - Vi) / a
= (0 - 3.5m/s) / 2.8 m/s^2
= 1.25s

The time for the skier to come to a stop is 1.25 seconds.

#### Doc Al

Mentor
inner08 said:
a) the time taken for the skier to come to a stop

For a:

Fa = Fgh + Ff
= Fg sin 5 (degrees) + ukFn
Realize that the kinetic friction opposes the skier's motion.

#### inner08

Oh sorry..i'm still trying to get the hang of all of this. Its been a while since I did some physics.

So since the kinetic friction opposes the skier's motion, I would use the following formula: Fa = Fg sin 5 (degrees) - ukFn? Then I would just plug in the values like I did right?

#### Doc Al

Mentor
Yes. Realize that the acceleration--given your choice of sign convention--must be negative: opposite to the skier's direction of motion. Otherwise, the skier would speed up instead of slow down.

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