(adsbygoogle = window.adsbygoogle || []).push({}); 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?

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# Homework Help: Forces - distance of a skier on an incline

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