# Homework Help: Skiing both ways conservation of energy/1D kinematic

1. Nov 16, 2013

### J827

skiing both ways....conservation of energy/1D kinematic

1. The problem statement, all variables and given/known data
A skier starts from rest at the top of a 45.0-m-high hill, skis down a 30° incline into a valley, and continues up a 40.0-m-high hill. The heights of both hills are measured from the valley floor. Assume that friction is negligible and ignore the effect of the ski poles.

a. How fast is the skier moving at the bottom of the valley?

b. What is the skier’s speed at the top of the second hill?

2. Relevant equations
Equation A
KEi + PEi = KEf + PEf

or

Equation B

3. The attempt at a solution
I know that the angle is irrelevant because only gravity is affecting his movement.

I understand part a. Using Equation A, initial KE and final PE are zero. Final velocity is 30m/s. I get the same answer using Equation B. I confirmed this answer with the solution manual.

I am getting stuck on part b. Here is what I have tried so far:

Option 1a:
starting from the valley, using Equation A
Initial PE = 0, because you are in the valley.
Mass factors out of equation because is in three remaining terms.

(0.5 * 302) = (0.5 * Vf2) + (9.8 * 40)
vf = 10.8 m/s

Option 1b:
same as 1a, but using -9.8 since he's going in a different direction
vf = 41 m/s

Option 2:
starting from the valley, using Equation B
vf2 = (302) + (2 * -9.8 * 40)
vf = 10.8 m/s

Option 3a (as taken from official solution manual):
starting from the top of the first hill, using Equation A
initial KE = 0
mass factors out of equation

ghi = 0.5vf2 + ghf
(9.8*45) = (0.5vf2) + (9.8 * 40)
vf = 9.9 m/s

Option 3b
(because I don't understand why the same sign for g would be used if he is going in 2 different directions)
(9.8 * 45) = (0.5vf2) + (-9.8 * 40)
vf = 41 m/s

2. Nov 16, 2013

### rude man

Your problem is roundoff error. You did not calculate your v at the bottom of the hill correctly. It's less than 30 m/s.

3. Nov 16, 2013

### J827

thanks.

4. Nov 16, 2013

### CWatters

Edit: looks like I took too long thinking about why g is positive. Anyway..

I make the velocity at the bottom 29.7m/s not 30m/s. Makes all the difference.

Otherwise ..

1a and 2 and 3a are all valid approaches and all give the same answer 9.9m/s

1b and 3b are wrong.

I've not got a good explanation for why g is positive accept to say that as he climbs the other hill he clearly must gain PE.

5. Nov 16, 2013

### rude man

g does not change - ever!

6. Nov 16, 2013

### haruspex

It all depends on how you define the positive direction. If up is positive in both parts then g is negative in both and the vertical distance is negative in a), positive in b).

7. Nov 16, 2013

### rude man

If I drop something it falls toward Earth. It doesn't matter if I'm going uphill or downhill.

8. Nov 16, 2013

### J827

while the magnitude of g never changes, how it affects your velocity will depending on how you have defined direction. if I am skiing downhill, my velocity will increase at a rate of 9.8 m/s2. As I ski uphill, however, my velocity will decrease at a rate of 9.8 m/s2.

that makes sense & explains why using when Equation A 9.8 is positive, while using Equation B 9.8 is negative.

9. Nov 16, 2013

### haruspex

Was that in response to my post? I don't see how what I said contradicts that. It's g either way, but depending on how you define the positive direction, g is either +9.8ms-2 or -9.8ms-2.

10. Nov 16, 2013

### rude man

I know you well enough to know that you know that the sign of g does not change once the coordinate system is defined!

No, I sensed the OP thought g was one sign going down & the other going up.

11. Nov 16, 2013

### J827

nope, I'm good. :)