- #1

- 26

- 0

http://www.physics.gatech.edu/people/faculty/larry/images/16212.gif

I keep getting mixed results.

I did 30(cos(40))*-1 to get the distance from A to B... but now what?

- Thread starter RedPhoenix
- Start date

- #1

- 26

- 0

http://www.physics.gatech.edu/people/faculty/larry/images/16212.gif

I keep getting mixed results.

I did 30(cos(40))*-1 to get the distance from A to B... but now what?

- #2

Doc Al

Mentor

- 44,953

- 1,220

I assume you mean 30*(1 - cos40). What's conserved?I did 30(cos(40))*-1 to get the distance from A to B... but now what?

- #3

- 26

- 0

I assume you mean 30*(1 - cos40). What's conserved?

Energy is conserved. But I honestly do not have a clue where to proceed. I am obviously missing the portion of this in the book. The answer is in front of me, but I am being blind.

- #4

Doc Al

Mentor

- 44,953

- 1,220

- #5

- 26

- 0

PE = mgy

KE = .5mv^2

so... 800N / 9.8 = 81.63

(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29/1000.. still wrong. where did I mess up?

- #6

Doc Al

Mentor

- 44,953

- 1,220

The left side is perfectly OK, but I'm not sure what you're doing on the right. Set the left side equal to .5mv^2 and solve for v.(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29/1000

- #7

- 26

- 0

I was converting to kN... it did not matter, I get it.The left side is perfectly OK, but I'm not sure what you're doing on the right. Set the left side equal to .5mv^2 and solve for v.

Bingo!

(.5 x 81.63 x 12^2) + (800 x 30(1-cos40)) = 11492.29

11492.29/.5/81.63 = 281.57 ; 281.57^1/2 = 16.78m/s

Excellent, thank you for your help!

- Last Post

- Replies
- 4

- Views
- 2K

- Replies
- 6

- Views
- 3K

- Last Post

- Replies
- 2

- Views
- 2K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 3

- Views
- 3K

- Last Post

- Replies
- 7

- Views
- 1K

- Last Post

- Replies
- 1

- Views
- 2K

- Last Post

- Replies
- 2

- Views
- 5K

- Last Post

- Replies
- 1

- Views
- 1K

- Last Post

- Replies
- 2

- Views
- 1K