# Can infinite work be done?

Wikipedia defines work as follows.

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

Also work is defined as the product of force and the displacement caused by the force. Now, if a block is kept on a frictionless surface and a force is exerted on it, it will travel indefinitely unless another unbalanced force acts on it.

Does this mean that the work done is infinite?

• Only if you can supply the force indefinitely. – ACuriousMind Oct 15 '15 at 16:12
• I still don't understand. If a force is exerted for a short period of time won't the object be displaced indefinitely? – The Cryptic Cat Oct 15 '15 at 16:16
• the work is computed only while the force is acting – user83548 Oct 15 '15 at 16:16
• note the words "when acting on a body" – garyp Oct 15 '15 at 16:38
• Say a force of $5$ N is applied on a body while it travels $10$m. Then the force is withdrawn but the body travels another $10$m due to inertia of motion. So is work done = $50$ J or $100$ J? – SchrodingersCat Oct 15 '15 at 18:35

## 4 Answers

Work is the result of a force on a point that moves through a distance. As the point moves, it follows a curve X, with a velocity v, at each instant. The small amount of work δW that occurs over an instant of time dt is calculated as

$\delta W = \mathbf{F}\cdot d\mathbf{s} = \mathbf{F}\cdot\mathbf{v}dt$

where the $F ⋅ v$ is the power over the instant $dt$. The sum of these small amounts of work over the trajectory of the point yields the work,

$W = \int_{t_1}^{t_2}\mathbf{F} \cdot \mathbf{v}dt = \int_{t_1}^{t_2}\mathbf{F} \cdot {\tfrac{d\mathbf{s}}{dt}}dt =\int_C \mathbf{F} \cdot d\mathbf{s}$

where $C$ is the trajectory from $x(t1)$ to $x(t2)$. This integral is computed along the trajectory of the particle, and is therefore said to be path dependent. If the force is always directed along this line, and the magnitude of the force is $F$, then this integral simplifies to

$W = \int_C F\,ds$

where $s$ is distance along the line. If $F$ is constant, in addition to being directed along the line, then the integral simplifies further to

$W = \int_C F\,ds = F\int_C ds = Fs$

where $s$ is the distance traveled by the point along the line.

In your case you do not say if the force is being applied constantly or not. We only know that is has been applied. Since we don't know if the force has stopped we have to assume at least two cases. The first, the force is being a applied constantly over an indefinite period of time on the block on a frictionless surface and in this case its velocity is accelerating over and infinite time period and then yes the work done that is approaching an infinite quantity. Or second in the case where the force may have been stopped being applied after a certain period of time the block would cease accelerating and continue at a constant velocity. In this case the force would have a definite quantity.

To understand this better lets examine the Work Energy Principle :

The principle of work and kinetic energy (also known as the work–energy principle) states that the work done by all forces acting on a particle (the work of the resultant force) equals the change in the kinetic energy of the particle. That is, the work $W$ done by the resultant force on a particle equals the change in the particle's kinetic energy $E_k$,

$W=\Delta E_k=\tfrac12mv_2^2-\tfrac12mv_1^2$

where $v_1$ and $v_2$ are the speeds of the particle before and after the work is done and $m$ is its mass.

The derivation of the work–energy principle begins with Newton's second law and the resultant force on a particle which includes forces applied to the particle and constraint forces imposed on its movement. Computation of the scalar product of the forces with the velocity of the particle evaluates the instantaneous power added to the system.

For additional edification -

In the general case of rectilinear motion, when the net force F is not constant in magnitude, but is constant in direction, and parallel to the velocity of the particle, the work must be integrated along the path of the particle:

$W = \int_{t_1}^{t_2} \mathbf{F}\cdot \mathbf{v}dt = \int_{t_1}^{t_2} F \,v dt = \int_{t_1}^{t_2} ma \,v dt = m \int_{t_1}^{t_2} v \,{dv \over dt}\,dt = m \int_{v_1}^{v_2} v\,dv = \tfrac12 m (v_2^2 - v_1^2)$

Let's Examine your first statement -

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

This true and for a point in time it is $\delta W$ as described above.

Next you say

"Also work is defined as the product of force and the displacement caused by the force. Now, if a block is kept on a frictionless surface and a force is exerted on it, it will travel indefinitely unless another unbalanced force acts on it."

OK but nothing is said about the time the force is applied. So as you can see we are left with answering a number of cases of time periods and also must cover a number of cases when the force is constant over time and when it is not but in the same direction. Hope the various methods of dealing with the potential cases in point have helped to show you the need for structuring your questions carefully.

And your question -

"Does this mean that the work done is infinite?"

Yes if the force is applied constantly or inconsistently over an infinite time period. No if it was only applied for a definite period of time.

Work is only being done on a body while it is accelerating. The work done is equal to the change in the kinetic energy of the body. If a body is moving at constant velocity its energy is not changing so no work is being done on it.

So in your example of the frictionless block, if we apply a force $F$ for a time $t$ then the work done will be given by $F\cdot x$, where $x$ is the distance the block moves in the time $t$ while the force $F$ is being applied. Once we've stopped applying the force it doesn't matter how much farther the block moves away, the force on it is zero and the block isn't accelerating so no work is being done on the block.

• I thought if the object is in motion and forces are being applied to it, work is being done. – TanMath Oct 16 '15 at 20:56
• do you mean "total work"? – user83548 Oct 16 '15 at 21:54
• @brucesmitherson no, wouldn't that be zero? – TanMath Oct 17 '15 at 3:30
• @brucesmitherson: yes, in this context work means net work i.e. total work done on the object minus the total work done by the object. – John Rennie Oct 19 '15 at 5:14
• @JohnRennie wrong answer and comment – user83548 Oct 23 '15 at 19:40

I will give the bounty to one the right answers other than mine, but I want to write one to correct the wrong concepts on the currently accepted answer, as it would be sad that the OP end up with the wrong concepts after using this site.

First affirmation: "Work is only being done on a body while it is accelerating. The work done is equal to the change in the kinetic energy of the body. If a body is moving at constant velocity its energy is not changing so no work is being done on it." Response: this is wrong, an example will be an object resting in the floor of an elevator that moves at constant speed. For an observer outside the elevator and at rest relative to the Earth there are two forces acting: gravity and the normal from the floor. Suppose the elevator moves upwards. Then the force by the normal force will be positive and the work by gravity negative (the force is opposite to the displacement). The object moves at constant speed and the total work done over the body is zero.

Second affirmation: "in this context work means net work i.e. total work done on the object minus the total work done by the object" Response: This is also wrong. the work done by the object is unrelated to the work done by the object. Using the same example as above, the object makes force on two other objects. First, the gravitational force that the object makes on Earth (the reaction to the force that the Earth does on the object), which makes only negligible work on Earth, as the Earth displacement is negligible. Second the reaction to the normal that the object makes on the elevator. This force makes negative work on the elevator as it points downwards and the elevator moves upwards. Thus, the total work done on the object is zero, and the total work done by the object is negative.

• Acceleration fields, like gravity, are a bit weird. Work done when moving vertically against gravity is non-zero. As the elevator moves upwards, it's doing positive work on the object inside. The other answers are talking about moving outside an acceleration field, or along an equipotential surface within the field, like the question is asking about. See this page for some more details. – MichaelS Oct 26 '15 at 1:17
• @MichaelS perhaps you should read again the reasons for the bounty – user83548 Oct 26 '15 at 1:20
• You should clarify precisely what you think is wrong with the answers in place. You just say the answers "are wrong". The post above is based on false presumptions. First, the question isn't asking about moving in a gravity field, so answers that don't discuss this point are incomplete, not wrong. Second, you are wrong in thinking no work is done by the elevator in your example. – MichaelS Oct 26 '15 at 1:32
• I agree that work done by an object isn't relevant to work done on an object, but that was in a comment, not part of the answer. And was likely intended to mean "net work on the object" rather than what it states. – MichaelS Oct 26 '15 at 1:36
• I said the elevator does work! (the normal force) – user83548 Oct 26 '15 at 1:46

When there is a change of the energy of an object, it is because some force is acting on it in a direction of its displacement. When all the forces in that direction stop or go to zero, there is no increase in energy, which is detectable through the absence of change of speed of the object or its position in some field of force. So when a force acts on a body for some finite period of time, it will increase its energy somewhat, but when it stops acting on it, the body continues to move at speed at which the force left it. Now the body moves with the constant energy. Why? Because there is no change in its speed. Energy of the body is related to the work done by force.