# Ramp problem

1. Oct 21, 2007

### Oliviam12

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

2. Relevant equations
Fg= -mg
F=MA
ect.

3. 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= .15 (-686)
Fk = -102.9 N

Net Force: -686.7 - -102.9= -583.8 N

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

Thanks!

2. Oct 21, 2007

### 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.

3. Oct 21, 2007

### learningphysics

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...

4. Oct 22, 2007

### Oliviam12

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

Last edited: Oct 22, 2007
5. Oct 22, 2007

### Oliviam12

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

6. Oct 22, 2007

### mgb_phys

For c, you have found the acceleration not the speed

7. Oct 22, 2007

### learningphysics

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

8. Oct 22, 2007

### 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.