# Potential Energy and Kinetic Energy Problem

1. Oct 5, 2007

### jayced

1. The problem statement, all variables and given/known data
Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
Vertical velocity up the mountain=0.0985m/s
Horizontal velocity down the mountain=0.141 m/s

A)Calculate the change in gravitational potential energy for Brandon and for Tom as they make their climb in the Rockies.
Delta GPE for Brandon=
Delta GPE for Tom=

B)Calculate the average kinetic energy for Brandon and for Tom as they climb up the mountain and as they descend.

KE for Brandon going up=
KE Tom going up
KE for Brandon coming down =
KE for Tom coming down=

2. Relevant equations
A)GPE=mgh
B)KE=1/2mv^2

3. The attempt at a solution
It seems to simple for question A is too simple to be true so I'm concerned about wether it is correct.
M computations:
GPE=mgh
=(91.1kg)(9.8m/s^2)h and (11.2kg)(9.8m/s^2)h

Question:What height should i use for h? Is this the right equation? What is meant by Change in gravitational energy,do I have to subtract them?

For B)
KE=1/2mv^2
KE for Brandon going up
KE=1/2(91.1kg)(0.0985 m/s^2)^2
=.441

KE Brandon going down
KE=1/2(91.1kg)(0.141 m/s^2)^2
KE=.905

Whats the units for KE?
Is this correct and if so should I do the same for Tom?

2. Oct 6, 2007

### learningphysics

change in gpe = $$mgh_2 - mgh_1 = mg(h_2-h_1)$$

so in your equation h refers to the change in height...

For the kinetic energy part, you need the net velocity... you used only the horizontal and vertical component of velocity in your formula.

"Vertical velocity up the mountain=0.0985m/s"

so what is the magnitude of the net velocity? use trig...

3. Oct 6, 2007

### jayced

Thank you Learn,

I dont understand because my avg velocity going up the mountain was 0.0985m/s. My avg velocity going down the mountain was 0.141m/s. But if I used my net velocity would that make my numbers the same for KE goin up and down for Brandon,becuase the only variable that changes is my mass. I'm really confused. Help clarify things for me please.

4. Oct 6, 2007

### learningphysics

This is what the question says:

Vertical velocity up the mountain=0.0985m/s
Horizontal velocity down the mountain=0.141 m/s

The average velocity going up the mountain is not 0.0985m/s... the average vertical velocity going up the mountain is 0.0985m/s... Brandon has both vertical and horizontal velocity... what is the horizontal velocity while Brandon is going up the mountain? What is the net velocity going up the mountain (vector sum of horizontal and vertical velocities) ?

5. Oct 6, 2007

### jayced

We are getting there Learn,

Can I find the horizontal velocity by doing my avg velocity of .705m/s times my time which is 22680s? Is my horizontal velocity equation v x t?

Then in order to find my net velocity I add my horizontal and vertical velocities,Correct?
But what does my net velocity solve for? I really need help with my forumulas because i am getting confused.
My avg velocity is .705m/s if that helps.

6. Oct 6, 2007

### learningphysics

How did you get 0.705?

7. Oct 6, 2007

### jayced

I got 0.705 when i divided the length of their path(16000m)/ the time it took them to get there 22680s,which equals 0.705m/s. That was their average speed in m/s.

8. Oct 6, 2007

### learningphysics

jayced, I'm confused by the question itself... is this just the question or does this include stuff you calculated:

"Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
Vertical velocity up the mountain=0.0985m/s
Horizontal velocity down the mountain=0.141 m/s"

so all of this is part of the question? they gave 0.0985m/s and 0.141m/s in the question?

9. Oct 6, 2007

### jayced

It was one problem but my teacher refernced it for more questions. For the first part I calculated:

A)What is the average angle of ascent the path makes with the horizontal?Theta=8 degre
sin theta =2235m/16000m the 2235 came from 2746m-511m=2235
sin-1(0.139)=8.03 degrees

B)What is their average speed along the path in meters/second? average speed=.705 m/s
Avg speed=d/t
So I took 16000m/22680s=.705m/s

C)What is their average vertical velocity? Vup=0.0985 m/s
DeltaV/t=Average velocity
2235m/22680s=0.0985 m/s

D)After some food they hike back down to their starting point in T2=4.24 hours
What is their average vertical velocity? Vdown=0.141 m/s
DeltaV/t=avg vertical velocity
2235m/15840s=0.141m/s

This is what I calculated right above. Did I clarify it enough? The only thing thats given was :

Brandon and Tom are climbing a mountain. He starts from an elevation of H1=511m above sea level. In 6.18 hours he climbs to an elevation of H2=2746m above sea level. The length of his path is L1=16000m.

Brandon has a mass M1= 91.1kg and Tom has a mass M2=11.2kg
Vertical velocity up the mountain=0.0985m/s
Horizontal velocity down the mountain=0.141 m/s

10. Oct 6, 2007

### learningphysics

ok. so the average velocity up the mountain is 16000/22680 = 0.705m/s. use this for the kinetic energy going up the mountain

How did you get 15840s for the time down the mountain? 4.24*3600 = 15264s

anyway for down the mountain, use 16000/time down the mountain... then use that velocity to get kinetic energy down the mountain.

11. Oct 6, 2007

### jayced

60mins=1hr 24mins= 24min*60s=1440s
60mins* 4hr=240mins
60s=1min
240min*60s=14400 +1440s= total seconds =15840

12. Oct 6, 2007

### learningphysics

Oh, I see... it's not 4.24 hrs, but 4hrs, 24 minutes... okay, so 16000/15840 = 1.01m/s. use this for the kinetic energy down the mountain...

13. Oct 6, 2007

### jayced

Wow ,thanx so much learningphysics you helped me plenty and this is the best kept secret. Good luck and know that its greatly appreciated.

14. Oct 6, 2007

### learningphysics

you're welcome.