# I dont understand the work equation

I don't understand how work = force * displacement as if a force of say 1 Newton was to be applied to two objects of different mass until the object reached a displacement of say 1 meter, surely the object of less mass would displace 1 meter in less time (due to faster acceleration) meaning the force would be applied for less time resulting in less work. I know there is something fundamentally wrong with my understanding of this but I'm not sure exactly what. any help would be greatly appreciated.

• Check that time does not play any role in work. It doesn't matter if you need 1 second or 1 year to move a stone, the work spent is the same – FGSUZ Jan 8 at 21:37
• @FGSUZ Don't you agree that time plays a role in the rate at which work is done? As I indicate in my answer, the OP is confusing the rate at which work is done with the amount of work done. – Bob D Jan 8 at 23:31
• Of course I agree. Work done is not "rate of work done", as you say. But I didn't mention the rate at all – FGSUZ Jan 8 at 23:34

One of the things that we can predict while applying work energy theorem is that the change in kinetic energy is the same for both the given case (faster as well as slower) given the fact that other forms of energy of the system change negligibly. Let's assume that these works aren't the same:

$$W=\Delta K$$ and $$W' = \Delta K'$$

[Note that mass doesn't show up in the above equation.]

Showing that time doesn't play a role (the way you think)

Let the same force $$F$$ act on two stationary objects of mass $$m$$ and $$m'$$ and there velocity becomes $$v$$ and $$v'$$ after some time $$t$$ and $$t'$$ respectively covering the same distance $$d$$.

$$a= \frac {F}{m}$$ $$a' = \frac {F}{m'}$$

Now since $$d = ut + \frac {1}{2}at^2$$

$$\Rightarrow d = \frac {1}{2}at^2$$ and $$d = \frac {1}{2}a't'^2$$

$$t = \sqrt {m\frac {2d}{F}}$$ and $$t' = \sqrt {m'\frac {2d}{F}}$$

$$\Rightarrow t'= t \sqrt {\frac {m'}{m}} \tag 1$$ Therefore $$v = t \frac {F}{m}$$ and $$v' = t' \frac {F}{m'}$$

Using eq. $$(1)$$ we get,

$$\Rightarrow v' =\frac {F}{ \sqrt {m m'}} t \tag {2}$$

Now the change in kinetic energy ($$\Delta K$$) is (clearly):

$$\Delta K = \frac {1}{2} mv^2 = \frac {F}{2m}t^2$$ and $$\Delta K' = \frac {1}{2} mv'^2$$ Now substituting $$v'$$ from eq. $$(2)$$ we get,

$$\Rightarrow \Delta K' = \frac {F}{2m}t^2$$ i.e.,

$$\Delta K = \Delta K'$$

$$\Rightarrow W = W'$$

This means that for same force acting for same distance the work done is the same.

Let’s take a little look at the definition of work.

The work done by a force in moving some system through a certain displacement is defined to be the force times displacelment.

Let’s take gravity as an example.

If you are holding a rock at a certain height and you let it fall, the gravitational force will be acting during the motion of the rock (it is by the existence of such force that the rock falls!)

You can imagine doing this experience in two different ways, one, in vacuum, and two, in the presence of an atmosphere. Of course there will be a different in the time the rock takes to reach the ground, because in the presence of an atmosphere you will have to account the air drag.

Although, gravity does the same amount of work in both cases, provided you let it fall at the same height and you are doing both experiences in the same place (so the acceleration due to gravity will be the same).

The work done by the force will be the same! Same force, same displacement.

For time, we can consider the impulse of the force which is defined as the force times the time interval (this time interval is the amount of time the force is acting on the system).

By that you can take the conclusions you stated in your question, except that the amount of work done is the same!

You are confusing the concept of "power" (that the user is giving to the system) with the concept of "energy" (that is something more inherent to the system).

Power is defined as $$P=\frac{dW}{dt}=\mathbf{F}\mathbf{v}$$ with $$\mathbf{v}$$ the velocity, $$\mathbf{F}$$ the force, and $$P$$ the power and $$W$$ the work done. Indeed the power you will need is going to be less for an item of a low mass than for an item of high mass.

You should also look at the concept of dissipation of energy. Indeed, the work $$W$$ will eventually be the same in the 2 cases, but the dissipation of energy is going to be higher in the second case.

The work that is done does not depend on time and is not connected to what power the user is giving to the system and what power is going to be dissipated by this system.

If you call $$R$$ the time derivative of the energy of the system, $$R=\frac{dE}{dt}+w$$, with $$E$$ the intern energy and $$w$$ the dissipation of energy.

I know there is something fundamentally wrong with my understanding of this but I'm not sure exactly what. any help would be greatly appreciated.

I think you are confusing the difference between the rate at which work is done, versus the amount of work done. You are thinking about the rate at which work is done, which is power.

As you know, work is force $$F$$ times displacement $$d$$. To make things simple, let the force be constant. Then

$$W=Fd$$

From Newton's second law

$$F=ma$$

So work is

$$W=(ma)d$$

Re writing Newton's second law

$$a=\frac{F}{m}$$

From the last equation you can see that if we halve the mass (m/2), for the same force $$F$$ we double the acceleration (2a). Putting these changes into the work equation we get

$$W=\biggr(\frac{m}{2}\biggl)(2a)d= (ma)d$$

So there is no difference in the work done if we halve the mass and double the acceleration as long as the force is the same. It will take less time to cover the distance, but the work done is the same.

Bottom line: If you reduce the mass for the same force over the same distance, you increase the rate at which work is done, but not the amount of work done.

Hope this helps.