Will the work done be infinite?

[Vijay Raghavan]

If I displace an object at rest in space by giving it a force F in X direction and the object tends to move forever, will the work done be infinite?
Knowing that work done = force*displacement . Since the space has no external resistance (unbalanced force) to stop the object from moving , making the displacement infinitely large.Let us assume that it stays clear of gravitational influences of other bodies in space.

 

[Dale]

Hi Vijay, welcome to PF!

The displacement is the displacement during the application of the force. So if you continue to apply the force forever then yes the work will be infinite. If, at some time, you stop applying the force then there will be no more work done, regardless of how long the object continues to drift with no force.

 

[Vijay Raghavan]

Thank you Dale , I am kind of new to this Forum .I am not sure how this forum works yet , so apologies for any inconvenience caused .Coming back to my question , This is what i understood in my high school about work done
Example when you push a wall of the room or when you stand in the room for a long time you feel exhausted and you think a lot of work is done, But in Physics the word Work Done has a definite meaning and is used in a restricted sense. Work is said to be done only if the point of application of force moves. For work do be done they must satisfy the following two criteria

1. Force should be applied
2. The Force produces displacement

So , irrespective of whether the force is instantaneous or continued , isn't work defined the same ?

 

[Dale]

Yes, work is defined the same. The place where you are going wrong is determining what displacement to use. The work is always calculated using the displacement during the application of the force.

I am not sure if you are familiar with calculus, but for a time-varying force it is easier to think in terms of power, which is the derivative of work. $P=f\cdot v = dW/dt$

 

[Vijay Raghavan]

Yes , I Am familiar with the basics of calculus. That does indeed clarify my doubt , thank you :)

 

[Monsterboy]

The displacement is the displacement during the application of the force. So if you continue to apply the force forever then yes the work will be infinite.

What if the force is applied only for an instant like an impact force where the two bodies are in contact only for an instant of time ?

 

[Vijay Raghavan]

What if the force is applied only for an instant like an impact force where the two bodies are in contact only for an instant of time ?

What Dale is trying to say is for time varying force like that where it is difficult to find the actual displacement (it might be just few x units, but thanks to Newtons law of inertia it will keep on moving ) because the object will move forever thanks to no other force acting on it . It is easier if we studied the objects power instead of work done. where dt is 0 . We know the force and we know the velocity.

 

[Monsterboy]

It is easier if we studied the objects power instead of work done. where dt is 0 . We know the force and we know the velocity.

Yes it is easier but I want the work only
So $P.dt$ = $dW$ if dt=0 then
$dW =0$

after integration

W= constant (not infinite)

How do you find the value of this constant ?
By work-energy principle
Work done = change in kinetic energy
Since the initial velocity is zero.
Work done W = (1/2)mv² ??

 

[PeroK]

Example when you push a wall of the room ... you feel exhausted and you think a lot of work is done,

You cannot easily tire yourself out by pushing on an immovable object. Certainly not compared with how quickly you tire if you move a large object.

 

[Vijay Raghavan]

Yes it is easier but I want the work only
So $P.dt$ = $dW$ if dt=0 then
$dW =0$

after integration

W= constant (not infinite)

How do you find the value of this constant ?
By work-energy principle
Work done = change in kinetic energy
Since the initial velocity is zero.
Work done W = (1/2)mv² ??

Yes it indeed looks that way , this was my thought initially where you use w=F*d when work is done against an opposing force otherwise the change in kinetic energy would be more appropriate.

 

[Vijay Raghavan]

You cannot easily tire yourself out by pushing on an immovable object. Certainly not compared with how quickly you tire if you move a large object.

I am not sure I get your point

 

[PeroK]

I am not sure I get your point

Try it. Push against a wall as hard as you can and see how long it takes your muscles to tire out. It would be a long time/

Then, if you can, try bench-pressing some heavy weight where you can only do 5-10 reps. Your muscles will be exhausted within 30 seconds.

 

[Vijay Raghavan]

Try it. Push against a wall as hard as you can and see how long it takes your muscles to tire out. It would be a long time/

Then, if you can, try bench-pressing some heavy weight where you can only do 5-10 reps. Your muscles will be exhausted within 30 seconds.

ha ha i got that ...what exactly is your point ? what has that got to do with my question asked? I just said what we speak of work in general is different from the work done in physics.

 

[PeroK]

ha ha i got that ...what exactly is your point ? what has that got to do with my question asked? I just said what we speak of work in general is different from the work done in physics.

You're the one who said that if you push against a wall you get exhausted. Reread your post #3.

I was just pointing out politely that you were wrong.

 

[Dale]

What if the force is applied only for an instant like an impact force where the two bodies are in contact only for an instant of time ?

If a finite force is applied for an instant then the displacement is indeed 0 and the work is 0. The only way to have a non zero work over an instant of time is to have an infinite force.

Then $W=F \cdot d= \infty 0$ which is undefined. So you must use some other method to find the work.