# Skier on a slope with friction (looking for final velocity)

#### Cmertin

1. The problem statement, all variables and given/known data
A 75Kg skier starts down a 50m high 10˚ slope. What is his speed at the bottom?
Part A: Consider skis frictionless
Part B: Assume that the coefficient of kinetic friction between the skis and the surface of the slope is µk=.05

2. Relevant equations
|Ff|=µk•|FN|
Fnet=-Ff+FN+Fg=0=ma
a=g•sin(ø) (pretend that chi is theta)
FN=ma•cos(ø)

Values I figured out from Part A:
The length of the slope (on the hypotenuse) is 288m
Initial velocity in both the X and Y is 0m/s
For part A, the final velocity is 31m/s (does not need to be checked)

3. The attempt at a solution
Part A:
I got part A with 31 m/s and I know it's right.

Part B:
0=Fnet=Fg+FN-Ff=ma (Newton's second law)
Ff=Fg+FN
FN=75Kg•9.81m/s2•cos(10˚)=724.6N
I think that I have to plug the normal force into the friction equation (|Ff|=µk•|FN|
) though I'm not sure if I have to and then I don't know what I would do after that.

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#### Redsummers

Part B:
0=Fnet=Fg+FN-Ff=ma (Newton's second law)
Ff=Fg+FN
FN=75Kg•9.81m/s2•cos(10˚)=724.6N
I think that I have to plug the normal force into the friction equation (|Ff|=µk•|FN|
) though I'm not sure if I have to and then I don't know what I would do after that.

Yeah, the friction force is the result of the normal force and the coeff. of kinetic friction product. Once knowing the friction force, you already have the two forces acting in the slope, don't you? One is the Fx (that should be a component of Fg) and the other one is the friction itself. Then you can work out the acceleration.

Edit: I don't see how do you say that Fnet = 0. If that was the case in the x axis, there would be no acceleration. Moreover, you state 'Ff=Fg+FN' how do you find it to be so? Again, friction force is the normal (or Fy) times the coeff. of kinetic friction. Drawing the FBD is really helpful.

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