Solving Skiing Problems: Net Force, Speed & Free-Body Diagrams

[Oliviam12]

Homework Statement

A skier goes down a slope with an angle of 35 degrees relative to the horizontal. Her
mass, including all equipment, is 70 kg. The coefficient of kinetic friction between her
skies and the snow is 0.15.
A. Please draw a free-body diagram of the skier.
B. Calculate the net force acting on the skier.
C. If the slope is 60 m long, what is her speed at the bottom of the slope,
assuming that she started from rest?

Homework Equations

Fg= -mg
F=MA
ect.

The Attempt at a Solution

Does this like correct? (Especially B and C, seeing as I have never done this type before)

My A is:
http://img88.imageshack.us/img88/1207/freend1.th.png

My B is:
Fg=-mg
Fg=-70(9.81)
Fg=-686.7 N

Fk= LaTeX graphic is being generated. Reload this page in a moment.Fg
Fk= .15 (-686)
Fk = -102.9 N

Net Force: -686.7 - -102.9= -583.8 NC.) (The length of the ramp dosn't really matter does it?)
F=MA
-583.8= 70A
-8.34 m/s^2 =A

Thanks!

 

[mgb_phys]

For part C use conservation of energy.
At the start she has a PE = m g h at the end this is all KE = 1/2 m V^2.
Work out what vertical distance she travels in going 60m at 35deg.

 

[learningphysics]

Your part b isn't right. have a look at this page:

http://www.ac.wwu.edu/~vawter/PhysicsNet/Topics/Dynamics/InclinePlanePhys.html

you can't add the gravitational force and friction the way you did... you need to add them as vectors... take 2 axes... one perpendicular to the incline...

The 3 forces in this situation are gravity, normal force and friction... gravity divides into 2 components... parallel to the incline, and perpendicular to the incline (that's what that link is about).

The forces parallel to the incline are: friction and parallel component of gravity. What do these add to?

The forces perpendicular to the incline are: normal force and perpendicular component of gravity. These add to zero because the skier isn't accelerating in our out of the incline...

 

[Oliviam12]

Sorry? I don't understand what to do? That site confuses me even more...

 

[Oliviam12]

I did it a different way; Fnet =fg(sin\theta-\mucos\theta) and got 392.668 N ? (For B part) and is C part correct?

 

[mgb_phys]

For c, you have found the acceleration not the speed

 

[learningphysics]

I did it a different way; Fnet =fg(sin\theta-\mucos\theta) and got 392.668 N ? (For B part) and is C part correct?

your formula looks right but I'm not getting that number... I get 309.18N

 

[learningphysics]

Part c... use conservation of energy, taking into account the work by friction...

Work by friction = final energy - initial energy

-\mu*mgcos(35)*25 = (1/2)mv^2 - mgh

or get the acceleration from the force in part b) divided by mass... then use kinematics.