- #1

nikechristo

- 2

- 0

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.

1. How much work must be done to double this velocity?

2. How much work must be done to halve the original velocity?

SOmeone please help me. :|

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 nikechristo
- Start date

- #1

nikechristo

- 2

- 0

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.

1. How much work must be done to double this velocity?

2. How much work must be done to halve the original velocity?

SOmeone please help me. :|

- #2

berkeman

Mentor

- 63,237

- 14,183

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.

1. How much work must be done to double this velocity?

2. How much work must be done to halve the original velocity?

SOmeone please help me. :|

Welcome to the PF, nikechristo. We do not do your work for you here on the PF. You must show your work before we can offer tutorial help. That's why there is a Homework Help posting template (which you deleted) that you should use to post your questions. There is a spot at the end to show your work so far.

So show us how you will approach these problems. What is the definition of work, in terms of forces and other things?

- #3

nikechristo

- 2

- 0

Work is done on an object whenever a force makes something move. This is the definition of work.

The general equation of work done is W = Force applied * distance.

A 40 Kg wagon is moving with a constant horizontal velocity of 10 m/s.

1. How much work must be done to double this velocity?

2. How much work must be done to halve the original velocity?

W = Force applied * distance

I don't get this question. I need the applied force or the force of friction to solve this question, rite? I don't think there is enough information to solve this, but i may be wrong. Is there a way to find the work done by using the mass and distance?

- #4

Snazzy

- 468

- 0

The work done is the change in kinetic energy.

Share:

- Replies
- 12

- Views
- 324

- Replies
- 8

- Views
- 392

- Replies
- 3

- Views
- 272

- Replies
- 6

- Views
- 573

- Last Post

- Replies
- 2

- Views
- 260

- Replies
- 1

- Views
- 275

- Last Post

- Replies
- 1

- Views
- 166

- Replies
- 29

- Views
- 686

- Replies
- 15

- Views
- 737

- Replies
- 21

- Views
- 1K