Can we say that work done can be used only for mechanical energy?

[Frigus]

When we say that we apply force to a wall then we didn't transferred energy because of no displacement, but I think we somehow increase many types of energy of the wall. For e.g:- potential energy of particles in wall e.t.c.

So according to me it means that we defined work only for mechanical energy (kinetic energy+potential energy)
But I can't conclude myself anything because physics don't work according to my reasoning...I can't find anything written about this query so can you please tell me whether I am wrong or right.

 

[PeroK]

When we say that we apply force to a wall then we didn't transferred energy because of no displacement, but I think we somehow increase many types of energy of the wall. For e.g:- potential energy of particles in wall e.t.c.

So according to me it means that we defined work only for mechanical energy (kinetic energy+potential energy)
But I can't conclude myself anything because physics don't work according to my reasoning...I can't find anything written about this query so can you please tell me whether I am wrong or right.

If you take a heavy object and lean it against a wall and leave it there, then it will apply a large force to the wall (and the wall will apply an equal force to the object). There will initially be some elastic potential energy stored near the point of contact, but you could leave the object there for years and there will be no work done by either object.

 

[Frigus]

If you take a heavy object and lean it against a wall and leave it there, then it will apply a large force to the wall (and the wall will apply an equal force to the object). There will initially be some elastic potential energy stored near the point of contact, but you could leave the object there for years and there will be no work done by either object.

But I think energy of wall is increased so energy is transferred by force(in elastic potential energy)and then even if we say that in many years no work will be done by both objects then this doesn't means that no work is earlier done by either of the object.

 

[Chestermiller]

What if the wall is an ideally rigid material that doesn't deform when a force is applied to it?

 

[Dale]

But I think energy of wall is increased so energy is transferred by force(in elastic potential energy)and then even if we say that in many years no work will be done by both objects then this doesn't means that no work is earlier done by either of the object.

The usual assumption is that the wall is rigid. So there is no energy transferred.

Work is not restricted to mechanical energy.

 

[Frigus]

What if the wall is an ideally rigid material that doesn't deform when a force is applied to it?

sorry for this silly question...
Thanks.

Hemant

 

[Frigus]

Please help me out to get out of this doubt,
We have derived this
V²-U²=2as--(I)
Equation from these 2
S=ut+$\frac {1}{ 2}$at²--(ii)
v=u+at--(iii)
From these I understand that difference between square of final and initial velocity depends on the displacement as given in equation (I)
Then we have another equation
F=ma--(iv)
Which we put in (I) to get this
$\Delta$K.E= $\vec F\cdot\vec d$
From all these things I understand that we can use this equation of work done for constant acceleration as from (I) from we can know how change of square of v_f and v_i from equation (I) depends upon distance so we will put our initial and final position and then we can know change in v_f² and v_i² then after some algebra we will get change in kinetic energy which is equal to force times displacement which we called work done.
Now my confusion is that how can we use this force times displacement in the situations which have constant velocity as if see carefully after some displacement(when their is only one force) velocity is increased so to further increase velocity by a unit we have to provide more energy than earlier as kinetic energy depends upon square of velocity. For example for Changing velocity from 50m/s to 51 m/s we of mass 1 kg have to provide energy of 101 joules but for same mass if we have to change kinetic energy from 1m/s to 2 m/s we have to provide energy of 3 Joules.
Now in case of constant velocity the work done by our force for same displacement should be less than the displacement in case of single force.

 

[Dale]

Now my confusion is that how can we use this force times displacement in the situations which have constant velocity

If you have constant velocity then $a=0$ so $F=ma=0$ and $W=F\cdot d=0$

For example for Changing velocity from 50m/s to 51 m/s we of mass 1 kg have to provide energy of 101 joules but for same mass if we have to change kinetic energy from 1m/s to 2 m/s we have to provide energy of 3 Joules.

Yes. This is correct.

Now in case of constant velocity the work done by our force for same displacement should be less than the displacement in case of single force.

I cannot understand what you are asking in this part. If it is important then can you rephrase it more clearly?

 

[Frigus]

If you have constant velocity then a=0a=0a=0 so F=ma=0F=ma=0F=ma=0 and W=F⋅d=0W=F⋅d=0W=F\cdot d=0

I was saying something different which I tried to explain in second point.

I cannot understand what you are asking in this part. If it is important then can you rephrase it more clearly?

Let's say 2 forces are acting on the object of same magnitude in opposite directions then net force is 0 but their is some constant velocity then after some time there will be some displacement due to velocity. Now if we want to find work done by individual force how can we calculate it because in this formula F.d the change in velocity square is included which will increase amount of energy transferred to object exponentially.

 

[Dale]

Let's say 2 forces are acting on the object of same magnitude in opposite directions then net force is 0 but their is some constant velocity then after some time there will be some displacement due to velocity. Now if we want to find work done by individual force how can we calculate it because in this formula F.d the change in velocity square is included which will increase amount of energy transferred to object exponentially.

Well, the energy does not increase exponentially, it is increases quadratically with respect to speed. However, at a constant speed (with some opposing force) that goes away and you are left with a transfer which is simply linear with respect to distance.

This is exactly the correct behavior. Suppose that you are pushing a block over a rough surface at constant speed, so the two forces are your pushing force and the friction force. The energy depends linearly on how far you push, so the power required is higher the faster you are pushing (linearly with respect to velocity). Intuitively that should make sense from your own experience pushing things along rough surfaces.

 

[anorlunda]

Let's say 2 forces are acting on the object of same magnitude in opposite directions then net force is 0 but their is some constant velocity then after some time there will be some displacement due to velocity. Now if we want to find work done by individual force how can we calculate it because in this formula F.d the change in velocity square is included which will increase amount of energy transferred to object exponentially.

You are confusing yourself with contradictions. In that one sentence you say "constant velocity" and "change in velocity". Are you confusing change in displacement with velocity?

In your example of two cancelling forces, the net force is 0 and the acceleration is zero, and the net work is 0. That is true no matter how big the two forces are as long as they add up to 0. So no, given those parameters you can not calculate the work done by each individual force; you don't know how big the forces are.

 

[Frigus]

net force is 0

I wanted to found work done by individual forces not net.
I was doing the question which I attached with this post then this query came to my mind.

 

[Frigus]

Well, the energy does not increase exponentially, it is increases quadratically with respect to speed. However, at a constant speed (with some opposing force) that goes away and you are left with a transfer which is simply linear with respect to distance.

This is exactly the correct behavior. Suppose that you are pushing a block over a rough surface at constant speed, so the two forces are your pushing force and the friction force. The energy depends linearly on how far you push, so the power required is higher the faster you are pushing (linearly with respect to velocity). Intuitively that should make sense from your own experience pushing things along rough surfaces.

Okay I agree ,
this was the thing I wanted to say. So then we can't use the formula F.d

 

[Dale]

So then we can't use the formula F.d

Certainly, we can. That is how we know that the work is linear with distance (and therefore the power is linear with speed).

 

[Frigus]

Certainly, we can. That is how we know that the work is linear with distance (and therefore the power is linear with speed).

Thanks!
Work is linear so that's why to increase the speed by a unit we have to cover distance in proportion to increase in kinetic energy because work is linear with distance.