# Are we doing work when we compress a ball?

Take a sponge ball and compress it. The net force acting on the body is zero and the body isn't displaced. So can we conclude that there is no work done on the ball?

• Then what is the exact definition of work is it not F. S? – Sai Charan Reddy May 22 '18 at 17:18
• One type of work is the application of strain while resisting a stiffness. The stiffness of low-density cellular solids such as sponges and foams may be low, but it is not zero. The work is transformed into strain energy stored within the material. – Chemomechanics May 22 '18 at 18:04
• $$\sum F\,d\ne\sum F\sum d.$$ – Yves Daoust May 23 '18 at 10:12
• Work is a scalar quantity even though it is the (scalar) product of two vectors... – Peter A. Schneider May 24 '18 at 14:06
• Work can be better defined as the force.(displacement of point of application of force) – Aditya Garg Dec 3 '18 at 14:21

According to Wikipedia:

...a force is said to do work if, when acting, there is a displacement of the point of application in the direction of the force.

In the case you described, we have two forces acting on the ball from the opposite sides and, according to the above definition and provided that the center of the ball does not move, both forces perform work.

It is obvious that the work is done on the ball, since it changes the shape and acquires some potential energy.

You can replace the sponge ball with a spring and see the same effect.

Before applying the force the spring has zero kinetic and spring potential energy.

If we apply the force slowly from both sides, then the spring does not accelerate, so the kinetic energy remains zero. But the compressed spring contains some non-zero potential energy. The work we have done has gone into increasing the potential energy of the spring.

$$(KE+SPE)_{after} - (KE+SPE)_{before} = W_{net}$$

The same thing happens in the squeeze ball.

the net force acting on the body is zero and body isn't displaced. So, can we conclude that there is no work done on the ball?

You can conclude that the kinetic energy of the ball hasn't increased. But the work can go into other forms of energy.

But the individual parts of the sponge ball are moving from their position.

However if you are holding a pre-compressed ball, and maintaing the compression, then no work is done.

Your argument might have been true if the ball was rigid and incompressible. But it's not.

• Whoops. Slip of words. – harshit54 May 22 '18 at 17:56
• But wait, you are probably thinking, I expend effort to hold the ball compressed! Aren't my muscles doing work? They are, internally, because the individual actin-myosin fiber bundles can't stay contracted (unless you are dead; that's what rigor mortis is); each one must continuously relax and then contract again, so there's microscopic back-and-forth motion, doing work, the energy of which is dissipated as heat. The overall muscle generates a constant force, because there are lots of fiber bundles and they take turns contracting. – zwol May 22 '18 at 20:42
• Contrast if you compress a ball between a heavy book and a table: the book does work on the ball until the ball's spring force balances gravity, and then everything stops moving and stays that way indefinitely. – zwol May 22 '18 at 20:47
• I don’t know much biology but this guy is asking about the work AS defined by the formula $F.S$ so he is not asking about work done by our muscles – harshit54 May 23 '18 at 6:27

This is a good question: calculating work for systems that aren't particles is tricky. Recall that if a particle moves with speed $$v$$ and experiences a force $$F$$, the rate of change of its kinetic energy is $$P = F v.$$ Now suppose we are considering a general system, such as a ball that can deform. Since the above logic holds for every individual particle of the system, the speed $$v$$ is the speed of the point of application of the force, while the power $$P$$ is the rate of change of the total energy of the ball, which includes internal energy, and the energy $$E_{\text{cm}} = M v_{\text{cm}}^2/2$$ due to the motion of the center of mass.

In your case, if you hold the ball and squeeze, you're applying forces to the ball at many points. At each point, the force applied is in the same direction as the velocity the point is moving, because they both point inward, so you are doing positive work on the ball. And indeed the ball's total energy is going up, mostly due to an increase in potential energy.

However, some textbooks will tell you the exact opposite: they would say that since the center of mass of the ball isn't moving, the work done is zero! This is wrong if we're considering work and energy in the usual sense, but correct if we instead consider the related notions of "center of mass (CoM) work" and "center of mass energy".

The idea behind CoM work is that the CoM of a system satisfies $$F = M a_{\text{cm}}$$ where $$F$$ is the total force on the system. This is formally the same equation as we would have for a single particle of mass $$m$$, so by the same proof of the work-kinetic energy theory for particles, $$P_{\text{cm}} = F v_{\text{cm}}, \quad P_{\text{cm}} = \frac{d E_{\text{cm}}}{dt}.$$ That is, the total force on the system times the velocity of the CoM gives the "CoM power", which is the rate of change of the kinetic energy associated with the motion of the CoM. In the case of the ball, the CoM is not moving, so the CoM work is zero. I stress though, that CoM work is not the same thing as work, it's just a formal tool that looks similar.

Unfortunately, many high school courses mix these two things up. If your course involves calculus, they might be consistent about using the ordinary work. But in an introductory course, teachers freely switch between ordinary work and CoM work, leading to inconsistencies.

To get used to the two ideas, here are some basic examples.

Example 1: accelerating bicycle. In an accelerating bicycle, the force that pushes the bicycle forward is the friction force with the ground. However, if the wheels are rolling without slipping, then at every moment, the relative velocity between the ground and the part of the wheel touching the ground is exactly zero. Thus the friction force performs zero work, so the energy of the bicycle-cyclist system is constant. This is because the increase in $$E_{\text{cm}}$$ is compensated by a decrease in the chemical energy of the cyclist. On the other hand, the friction force does perform CoM work, and indeed $$E_{\text{cm}}$$ is increasing.

Example 2: walking up stairs. As a person walks up the stairs, the relative velocity between the foot touching the stairs and the stairs is zero. Thus the normal force from the stairs performs zero work, so the energy of the person is constant. This is okay, because the increase in gravitational potential energy is compensated by a decrease in the chemical energy of the person. On the other hand, the normal force does perform positive CoM work, which cancels out the negative CoM work due to gravity.

I think what is relevant is that forces and displacements are both vector quantities. Though the net force on the ball may be zero, when you calculate the work done when displacing it through each (say) $\bf{\delta x}$ is $\bf{F . \delta x}$ and around the ball all the dot products of the force $\bf F$ with each $\bf \delta x$ takes the same sign.

If you do not want the centre of mass of the ball to move you must apply two equal magnitude opposite direction forces on the ball with the line of action of the forces passing through the centre of mass of the ball.
Certainly the net force on the ball is zero and that results in the kinetic energy of the centre of mass of the ball being unchanged during the compression.

When the ball is squashed by these forces the displacement of each force is in the same direction of each force so each of the forces does work on the ball which results on the ball being compressed.
When the compression of the ball has stopped the applied forces stop moving and so no further work is being done by the applied forces on the ball.

So what happens to the ball as it is being compressed and work is done by the applied forces?
This depends on the material of which the ball is made and the elastic or non-elastic properties of the ball.

At one extreme when the applied forces are removed the ball stays compressed.
This is equivalent to an inelastic “collision” and the work which was been done on the ball when it was compressed increased its temperature and also results in molecular bonds within the ball being permanently broken.

At the other extreme all the work done increases the elastic potential energy of the ball which is equivalent to an “elastic” collision.
When the applied forces are removed the elastic potential energy of the ball would result in the ball expanding to its original shape gaining kinetic energy and then expanding further with the expansion eventually stopping and with the ball again having elastic potential energy.
So the shape of the ball would oscillate with there being some form of damping (ie not quite perfectly elastic) causing the ball to finally stop oscillating and become hotter.

When the applied forces are removed your sponge ball would probably expand towards its original shape and get hotter with the rise in temperature being too small to measure.
The increase in temperature would be noticeable if you kept repeated compressing and releasing something like a squash ball.

The rise in temperature indicates that the internal energy of the ball has increased and that energy must have come from the work done by the forces which were applied to the ball.

You do work to squeeze the ball, and then the ball does work to re-shape itself. How does an inanimate ball do work? It uses the energy you stored in it by compressing it.

You are correct that the net work is zero between the point where you started squeezing it and the point where it finished reforming. But net work is like net displacement. Something can move within a timeframe but still have zero net displacement. You did work but it was canceled out soon after. Imagine you take the ball and squish it, then stuff it into a bag too small for it to expand again. Now the net work is non-zero, because the ball didn't get to do work. It's applying a force to the bag, but can't apply it over distance because the bag is too strong. Anyone who has stuffed a puffy sleeping bag or a tent back into it's travel bag can tell you they did not come away from the experience having done no work.

The concept of work is contextual in that, in order to be useful, you need to define what it applies to. The easiest way to see this is when you calculate efficiency which is defined as work in divided by work out. We know that all the energy we put into something must go somewhere or it would violate the laws of thermodynamics. This, therefor is a pointless calculation unless you narrow down what you mean by work out.

For example, if we want to calculate the efficiency of a vehicle, we calculate the work put into it (i.e. fuel consumed) and the work produced which would typically be the distance the payload was moved. That the motor produced heat that was used to warm the cabin may or may not be relevant to that calculation depending on what you mean to evaluate.

So in this case, if you define productive work to mean that the ball was moved to a new location, then you can say no, no work was produced. However work was put in and therefore the system is perfectly inefficient. If you want to evaluate the efficiency of a squeezing machine like a juicer, you would define productive work (work out) differently and get a different result.

## protected by Qmechanic♦May 22 '18 at 18:11

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