- #1

TyErd

- 299

- 0

a. What is the magnitude of the force of friction between the block and the floor?

b. What is the maximum kintetic energy of the block?

You are using an out of date browser. It may not display this or other websites correctly.

You should upgrade or use an alternative browser.

You should upgrade or use an alternative browser.

- Thread starter TyErd
- Start date

- #1

TyErd

- 299

- 0

a. What is the magnitude of the force of friction between the block and the floor?

b. What is the maximum kintetic energy of the block?

- #2

joriarty

- 62

- 0

We won't do your homework for you. We'll only help you understand how to tackle the problem.

- #3

TyErd

- 299

- 0

- #4

joriarty

- 62

- 0

Have you been going over coefficients of kinetic friction recently in class/lectures?

- #5

joriarty

- 62

- 0

- #6

TyErd

- 299

- 0

yes your interpration is correct, it doesn't lift off the ground.

- #7

joriarty

- 62

- 0

- #8

TyErd

- 299

- 0

firstly, how do i figure out how much kinetic energy lost due to friction?

- #9

joriarty

- 62

- 0

At your level of study you can safely assume that all of this energy is transferred to the block

If this information still does not help you finish the first question, that would indicate you do not understand the fundamental physics behind the problem. In which case you should revise your notes and your textbook or consult your teacher/lecturer - there's no point attempting problems if you do not understand the background theory. We don't have all day to help you and I'm not going to just hand you the answers.

- #10

TyErd

- 299

- 0

I understand the kinetic energy parts, it is friction that i am having trouble with.

- #11

joriarty

- 62

- 0

- #12

TyErd

- 299

- 0

W=Fx

- #13

TyErd

- 299

- 0

- #14

TyErd

- 299

- 0

problem is how to calculate energy loss due to friction.

- #15

joriarty

- 62

- 0

I am presuming that you may assume that over the 1m distance where the block is still in contact with the catapult, the force exerted on the block is constant.

I have to go now and won't be able to check back for a day or two so you're probably on your own from now unless someone else helps you - but you're almost there. Good luck.

- #16

TyErd

- 299

- 0

ok thnx for the help

- #17

TyErd

- 299

- 0

any1 else want to help me out?

Share:

- Replies
- 12

- Views
- 584

- Replies
- 11

- Views
- 517

- Replies
- 10

- Views
- 955

- Last Post

- Replies
- 5

- Views
- 836

- Last Post

- Replies
- 11

- Views
- 168

- Last Post

- Replies
- 6

- Views
- 154

- Last Post

- Replies
- 17

- Views
- 399

- Last Post

- Replies
- 2

- Views
- 399

- Last Post

- Replies
- 2

- Views
- 794

- Last Post

- Replies
- 9

- Views
- 313