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 !

 

[rwisz]

I can see that you have made some good attempts at solving this problem. However, there are a few things that need to be corrected in your solution. Firstly, the horizontal component of gravity should be calculated as Fgx = mgsin4.7, where m is the mass of the skier and g is the acceleration due to gravity (9.8 m/s^2). Similarly, the normal force should be calculated as Fgy = mgcos4.7.

Additionally, the net force equation you have used is incorrect. The correct equation is Fnet = Fapplied - Ffriction, where Ffriction is the sum of the kinetic friction force and the force of gravity in the direction of motion (Fgx).

Using the correct equations, we can calculate the net force as:
Fnet = (mgsin4.7) - (0.11mgcos4.7) = (m)(9.8)(sin4.7 - 0.11cos4.7)

We also know that the net force is equal to ma, so we can equate the two equations and solve for a:
ma = (m)(9.8)(sin4.7 - 0.11cos4.7)
a = 9.8(sin4.7 - 0.11cos4.7)

Now, we can use this value of acceleration to calculate the distance traveled by the skier:
(v final)^2 = (v initial)^2 +2*a*d
0 = (2.7)^2 + 2(9.8)(sin4.7 - 0.11cos4.7)(d)
d = 13.03 m

This is the correct distance traveled by the skier before coming to rest. It is important to carefully consider all the forces acting on the skier and use the correct equations to solve the problem. Keep up the good work!