Forces - distance of a skier on an incline

[shawli]

Homework Statement

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.

Homework 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

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?

 

[mizzy]

Homework Statement

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.

Homework 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

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?

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

 

[bp_psy]

1.
So, solving, I get:

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

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

 

[shawli]

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

Thank you !