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

  • Thread starter damndamnboi
  • Start date
In summary, according to the physics textbook, work done is equal to force multiplied by the distance moved by an object in the direction of the force. However, in a scenario with no friction force, this equation can lead to ridiculous results. For instance, in an example with an object of mass 5kg, no friction or gravitational force, and a constant force of 10N applied for 5 seconds, the object will continue to move at a constant velocity of 10ms-1 even after the force is removed. This means that the distance moved by the object will be infinite. However, this does not mean that the work done is infinite, as the distance variable only refers to the distance over which the force is applied, not the
  • #1
damndamnboi
5
0
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 gravitational 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:
Physics news on Phys.org
  • #2
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
G01 said:
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.

I see. I think it would be better if we use the change in K.E to calculate the workdone in this scenario
 
  • #4
The correct formula for work is

[tex]W=\oint_C{\vec{F}\cdot d\vec{r}}[/tex]

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

[tex]W=F\times d[/tex]

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

[tex]W=\Delta KE[/tex]

Where KE is the kinetic energy, so in your example we can plainly see the work done is 250J.
 
  • #5
damndamnboi said:
I see. I think it would be better if we use the change in K.E to calculate the workdone in this scenario

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
damndamnboi said:
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 gravitational 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?

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
ZapperZ said:
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?
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
Workdone=force x distance
The original poster didn't know that the S was the distance over which the force was applied

What distance means ?
How can be measured in the scenario that the OP presents:
For example, an object of mass 5kg, in space with no friction force or gravitational force

From the object of mass 5 Kg to ...
 
  • #9
mgb_phys said:
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'.

Come again?

Here was the original post.

For example, an object of mass 5kg, in space with no friction force or gravitational 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?

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
Go1 answered the question.

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
In space with no friction force or gravitational 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 don't know why nobody replies.
 
  • #12
I good way, when you first deal with energy, is to think of it like this:

[tex]E = F(x_2 - x_1)[/tex]

Where [tex]x_1[/tex] is the point at which the force is applied and [tex]x_2[/tex] 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
GoldPheonix:
I good way
I don't understand

GoldPheonix:
E = F(x2-x1)

what I am asking is what "x" mean. ( please read my last post in this thread )

OP:
According to my physics textbook: Workdone=force x distance

The physics textbook doesn't explain what distance is ?

Most of you know the answer, but you haven't learned it from the physics textbooks. I am waiting your posts.
 

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

1. What is the formula for W=FXS?

The formula W=FXS represents the work done (W) on an object, which is equal to the force (F) applied to the object multiplied by the distance (S) over which the force is applied.

2. What does the "S" in W=FXS refer to?

The "S" in W=FXS refers to the distance over which the force is applied. It can also be thought of as the displacement of the object.

3. Is the distance in W=FXS a scalar or vector quantity?

The distance in W=FXS is a scalar quantity, meaning it only has magnitude and no direction. This is because it is a one-dimensional measurement along the direction of the force.

4. Can the distance in W=FXS be negative?

Yes, the distance in W=FXS can be negative. This indicates that the force is applied in the opposite direction of the displacement, resulting in negative work being done on the object.

5. How is the distance in W=FXS different from displacement?

The distance in W=FXS refers to the total length of the path over which the force is applied, while displacement refers to the straight-line distance between the initial and final positions of the object. Displacement is a vector quantity, while distance is a scalar quantity.

Similar threads

Replies
21
Views
3K
Replies
4
Views
15K
Replies
2
Views
1K
Replies
1
Views
964
Replies
2
Views
1K
Replies
16
Views
1K
Replies
5
Views
2K
Replies
1
Views
3K
Back
Top