# Work question without distance, work and velocity

1. Dec 8, 2011

### MsHomework

1. The problem statement, all variables and given/known data

How much work is done on an 8.0kg wagon rolling along a flat sidewalk, if applied force of 60N opposite to its direction of motion brings it to a rest in 2.0s?

2. Relevant equations
W = F · d

3. The attempt at a solution
I don't know how to start!

2. Dec 8, 2011

### MrB8rPhysics

Force is the rate of change of momentum... F = (mv-mu)/t

The mass is constant, and the wagon comes to rest.

This should allow you to find the initial velocity of the wagon.

From there either use an equation of motion to find how far it goes, then apply W = f.d, or calculate the kinetic energy of the wagon at its initial velocity - the work done to bring it to rest is equal to that initial KE.

3. Dec 8, 2011

### MsHomework

What does F= (mv-mu)/t stand for? Also the equation for motions would be the five main equations, right?

4. Dec 8, 2011

### MrB8rPhysics

F=(mv-mu)/t is a form of Newtons second law. m for mass, t for time, F for force, v for final velocity, u for initial velocity. Your F here is a force opposing motion so can be used in this formula as -60 N.

Five main equations? I teach them as the 4 SUVAT equations, if that rings a bell...

5. Dec 8, 2011

### e.bar.goum

F = force
m = mass
v = initial velocity
u = final velocity
t = time

It's just the change in momentum caused by stopping the cart.

I'm not sure what you mean by "five main equations", but an equation of motion is simply an equation that relates velocity and/or position and/or acceleration and time. Say, s = ut + 1/2 at^2

EDIT - Snap! Sorry.

6. Dec 8, 2011

### MsHomework

d=(u+v/2)t
u=v+at
d= vt+1/2at^2
d=ut-1/2at^2

7. Dec 8, 2011

### netgypsy

If you make a list of your given variables it makes it a lot easier to see which equation is the simplest to use.

You have force, mass, initial velocity and time. You can go with change in kinetic energy equations or find your distance and use your work equation. Several ways to work it.

8. Dec 8, 2011

### MsHomework

I get a negative intial velocity, is that normal?

9. Dec 8, 2011

### e.bar.goum

Depends on how you set up your problem. Negative values indicate the object is moving in the opposite direction to the direction you set positive to be. (or you made a mistake :D )

10. Dec 8, 2011

### MsHomework

Like I did F= (mv-mu)/t and my initial velocity was -15 m/s

11. Dec 8, 2011

### netgypsy

If you drew your force and velocity vectors what do you notice about their directions? Whether your initial velocity is negative depends totally on which direction you chose as positive, You could be correct or incorrect.

12. Dec 8, 2011

### MsHomework

so the intial velocity is in the opposite direction it is heading?

13. Dec 8, 2011

### MsHomework

Can someone just show me this step by step....because I am lost! This way I could try a similar question to practice this method!!!!

14. Dec 8, 2011

### I like Serena

Hi MsHomework!

Let's take a new approach from scratch.

Suppose the wagon was initially at rest and is accelerated by a force of 60 N for 2 seconds.
First question: what would the acceleration be?

15. Dec 8, 2011

### MsHomework

a= -7.5 m/s^2

16. Dec 8, 2011

### MsHomework

But isn' the velocity final at rest since it comes to a rest...?

17. Dec 8, 2011

### I like Serena

Yes.
I assume you used the formula F=ma to find this?

However, since I was talking about accelerating from rest, let's take the acceleration to be positive shall we?

Next question: what would the speed be after 2 seconds?

18. Dec 8, 2011

### I like Serena

Well, I've changed the setting of the problem.
In your problem statement, the wagon is moving along at an unknown speed and is decelerated to a halt.

But, if you'll bear with me, I first want to do a simpler problem, which is the reverse.
That is, we start from rest and then accelerate to the unknown speed.

19. Dec 8, 2011

### MsHomework

I don't have distance!

20. Dec 8, 2011

### MsHomework

alright I will follow your way.....so What do i do for distance???