# W=FXS, what distance does the S refers to?

1. Jul 18, 2007

### damndamnboi

According to my physics textbook: Workdone=force x distance moved by object in the direction of the force. This is ridiculous in a scenario with no friction force.

For example, an object of mass 5kg, in space with no friction force or gravitional force is accelerated from rest to a velocity of 10ms-1 by a force of 10N over 5seconds. From newton first law, the object will continue to move at a velocity of 10ms-1 even after the force applied is removed and will continue forever unless another force is applied to stop it. As such, the distance moved by the object in such a scenario will be infinite. Infinite x 10N = infinity. Does this means a force 10N applied for 5s can transfer infinite amount of energy?

Last edited: Jul 18, 2007
2. Jul 18, 2007

### G01

S refers only to the distance over which the force acts, not the total distance traveled. The object only moves a finite distance while the force is actually acting on it.

3. Jul 18, 2007

### damndamnboi

I see. I think it would be better if we use the change in K.E to calculate the workdone in this scenario

4. Jul 18, 2007

### nicktacik

The correct formula for work is

$$W=\oint_C{\vec{F}\cdot d\vec{r}}$$

You will probably need vector calculus to understand this. The generalization of a 1-dimensional constant force is

$$W=F\times d$$

In your example, the work done is not infinite, since the force is only applied for 5 seconds. The distance variable means the distance for which the force is applied. Once the force ceases to be applied, no more work is done!

Another way to calculate work is

$$W=\Delta KE$$

Where KE is the kinetic energy, so in your example we can plainly see the work done is 250J.

5. Jul 18, 2007

### G01

It is since you know the final speed and not the distance the force acts over. On the other hand, you may be in the exact opposite situation, in which case W = F*s is much more helpful. So, yes, it does depend on the scenario.

6. Jul 19, 2007

### ZapperZ

Staff Emeritus
I don't know what the problem is here. In the latter part, the force required to move the object is zero. Therefore, work done is 0 no matter what the displacement is since force is zero. So what's the problem?

Zz.

7. Jul 19, 2007

### mgb_phys

The original poster didn't know that the S was the distance over which the force was applied not the total distance moved.

I had a similair question from a student who didn't understand why a car moving at constant into a wall had any force because there was no 'accelaration'.

8. Jul 19, 2007

### alvaros

What distance means ?
How can be measured in the scenario that the OP presents:
From the object of mass 5 Kg to ...

9. Jul 19, 2007

### ZapperZ

Staff Emeritus
Come again?

Here was the original post.

I was referring to the latter, i.e. after "the force applied is removed". This means that the object is moving with a constant velocity. It takes zero force to do this, and so I asked what's infinite? The force applied is zero, and no matter what the displacement is, the work done is zero. That should have been the end of that story. But somehow, the OP still kept the force being applied for this very long distance (notice he/she had the "x 10N" part in the work equation), and equate that with the initial part of the dynamics, which of course makes no sense.

Zz.

10. Jul 19, 2007

### mgb_phys

The original poster was confused because they knew W=FS but wondered how if the object moved an infinite distance did you not have an infinite amount of W. This is of course because the S is only the distance the force is moved through - a typical case of knowing the equation but not knowing how to apply it.
Your point is a fuller statement of this, there are two regimes a large force*a short distance followed by a zero force for an infiite distance.

My point about the car wasn't really relevant to the question - it was just another example where the student knew the equation (f=ma) but didn't apply it properly because they didn't understand what the accelaration was.

11. Jul 20, 2007

### alvaros

In space with no friction force or gravitional force you push a stone direction East, your body is acelerated direction West.

The distance is from the stone to: your body, the center of gravity, the stars..

I think this is important, I dont know why nobody replies.

12. Jul 20, 2007

### GoldPheonix

I good way, when you first deal with energy, is to think of it like this:

$$E = F(x_2 - x_1)$$

Where $$x_1$$ is the point at which the force is applied and $$x_2$$ is the point there is stops.

(This equation only works such that the force is constant in one direction, you'd need calculus to resolve things diverging from this scenario)

13. Jul 21, 2007

### alvaros

GoldPheonix:
I dont understand

GoldPheonix: