Forces - distance of a skier on an incline

1. Jul 11, 2010

shawli

1. The problem statement, all variables and given/known data

A skier on a slope inclined at 4.7 degrees to the horizontal pushes on ski poles and starts down the slope. The initial speed is 2.7 m/s. The coefficient of kinetic friction between skis and snow is 0.11. Determine how far the skier will slide before coming to rest.

2. Relevant equations

uK= 0.11

v initial= 2.7 m/s
v final= 0.0 m/s

Fk = uK*Fnormal

F=ma

(v final)^2 = (v initial)^2 +2*a*d

3. The attempt at a solution

(Refer to system diagram for notations and stuff!)

So from what I can tell, the applied force is the horizontal component of gravity, Fgx, and the normal force is the vertical component of gravity, Fgy.

The horizontal component of gravity is:
Fgx= (sin4.7)(9.8)(m)

The force of kinetic friction would come out to be:
Fk=uK*Fgy
Fk=(0.11)(cos4.7)(9.8)(m)

The net force would be:
Fnet = Fapplied - Fkinetic
Fnet = Fgx - Fk
Fgx - Fk = m*a

So, solving, I get:

(sin4.7)(9.8)(m) - (0.11)(cos4.7)(9.8)(m) = m*a
a = 8.67 m/s^2

Now I can use acceleration to find the distance traveled by the skier:

(v final)^2 = (v initial)^2 +2*a*d
0 = (2.7)^2 +(2)(8.67)(d)
d= 0.42 m

The actual answer is 13m and I'm sure I've done something wrong .
Can someone guide me through this, please?

Attached Files:

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2. Jul 11, 2010

mizzy

CHECK YOUR CALCULATION FOR ACCELERATION. YOU SHOULD BE GETTING -0.27m/s^2.

3. Jul 11, 2010

bp_psy

You have some calculation error here. Be sure to do you calculations using degrees not radians.

4. Jul 11, 2010

shawli

Oooh, whoops. So my method was right all along?

Thank you !