Suppose we have a body initially at rest. Now a force ($F$) is continuously applied on it and it gets displaced by some distance $x$.

My tutor said that from work energy theorem it gains kinetic energy equals to $Fx$ and this work is the cause of gain in kinetic energy.

But I have some confusion:

If this work comes earlier to kinetic energy and is the cause of gain in kinetic energy then why did the body moved in the first place if no motion is possible if velocity of a body is zero?

Let me clarify:

Earlier the body was at rest. When force was applied, it gains some velocity and thus gains kinetic energy and after gaining velocity, the body moves or displaces.

So doesn't that mean that the body gained kinetic energy first and then work was done?

More importantly, which of the two (work and kinetic energy gain) occurs first?

  • 2
    $\begingroup$ I've removed a number of comments that should have been answers, and discussion about them. $\endgroup$
    – rob
    Commented Jan 15, 2021 at 22:41
  • 4
    $\begingroup$ Why does anything have to "come first" at all? Equations are equations, not causal relationships. Physics does not have a formal concept of causation. $\endgroup$
    – ACuriousMind
    Commented Aug 20, 2021 at 11:47
  • $\begingroup$ (Updated to correct an equation error). There is no requirement that things happen sequentially. For the application of force where work is done, W=ΔKE=Fd=(ma)d, and work, change in kinetic energy, and acceleration, ALL happen simultaneously. $\endgroup$ Commented Aug 20, 2021 at 23:39

7 Answers 7


I think that the discussion related to some answers and the original question revolves around the proper definition of causality in physical theories (I won't touch here the philosophical issues connected to this concept).

In physics, we say that a time-dependent quantity $A(t)$ has a causal relation with another time-dependent quantity $B(t)$, if we have a theory connecting $A(t)$ to $B(t')$ for all the times $t>t'$. In such a case, we say that $A$ depends causally on $B$.

The temporal sequence between cause and effect has a central role. This implies that a relation at equal times, $A(t)=B(t)$ can never be seen as causal.

Before getting the work-energy theorem, let me discuss some other examples of causal and not-causal relations in classical mechanics as a preliminary exercise.

The Second Law of Newton's dynamics is not a causal relationship between force ($F(t)$) and the acceleration $a(t)$ of a body subject to that force. It is just a formula connecting these two different quantities at the same time. The same holds (notwithstanding some widespread misconception) for all the action-reaction pairs of forces of the Third Newton's law. Again the fact that force $F_{12}(t)$ of the body $2$ on body $1$ is always equal and opposite to the force $F_{21}(t)$ of the body $1$ on body $2$, at the same time, does not imply a causal relation. In Newtonian dynamics, it is a property of the system of two bodies valid at each time $t$ in a completely symmetric way.

The Second Law of Newton's dynamics implies a causal relation between the velocity (or position) at time $t$ and the acceleration (then the force) at a previous time $t'<t$. This is formally evident by writing the solution of the differential equation for the velocity $v(t)$ as $$ v(t)= v(t') + \int_{t'}^t a(\tau) {\mathrm d} \tau. $$ I find it even clearer a discretized version of the previous equation: $$ v(t+\Delta t) = v(t) + a(t) \Delta t + O(\Delta t^2). $$

Now the discussion of the work-energy theorem in terms of causality should be simple.

The theorem states that the work done from time $t_0$ to time $t$ by the resulting force ${\bf F}$ along the trajectory followed by a particle of mass $m$, moving under the action of that force, is equal, at each time t, to the difference of kinetic energy at time $t$ and time $t_0$: $$ W_{t_0}(t)=\int_{t_0}^t {\bf F}(\tau)\cdot {\bf v}(\tau) {\mathrm d}\tau=\int_{t_0}^t m \frac{d{\bf v}(\tau)}{dt}\cdot {\bf v}(\tau) {\mathrm d}\tau=\frac12 m \left( v^2(t)- v^2(t_0)\right)=\Delta K_{t_0}(t). $$ It is clear that, for any choice of the initial time, we have the equality of two functions of the time at the same time. Therefore, based on the previous discussion, we do not have a causal connection between $K_{t_0}(t)$ and $W_{t_0}(t)$.

  • $\begingroup$ Interesting post. Work is force integrated along displacements. Kinetic energy contains a dot product of velocities. Owing to the causal nature of ${\bf a}$ with ${\bf v}$ and ultimately ${\bf r}$, then doesn't the velocity come before the displacement? The increment in work is $dW = {\bf F} \cdot d {\bf r}$ while the increment in kinetic energy is $dK = m {\bf v} \cdot d {\bf v} $. So doesn't $dK$ happen before $dW$? $\endgroup$
    – Evan
    Commented Aug 20, 2021 at 17:46
  • $\begingroup$ But they are both $d$something! You can not temporally order differences (albeit infinitesimal). At best, you can write something like (in 1D with constant F) $dW=Fdx=m{dx\over dt}dv$ solve the differential equation and get $v(t)-v_0={F\over m}dt=at$ which brings us again to "what comes first, the change in speed or the acceleration?". They are just the same thing, they are equal, if you measure both in the laboratory you get the same thing, but you can't predict one from the other, because the moment you have one... you have both. Cause or effect at same time is just a personal perspective. $\endgroup$
    – JalfredP
    Commented Aug 20, 2021 at 18:06
  • $\begingroup$ @GiorgioP thanks 😊 for your lovely answer. But one thing to know :- in F=ma , is it physically wrong to say that the force caused the acceleration ? $\endgroup$
    – Ankit
    Commented Aug 20, 2021 at 18:24
  • $\begingroup$ @Ankit In the strict sense of a cause-effect relation, yes, it is not correct. A more accurate description is that the force and initial conditions are the causes of a specific motion. It may be easier to see this point by considering the inverse problem. If we know the acceleration at the time $t$, and the mass, we can obtain the total force at the same time. Cause and effect exchanging their role? $\endgroup$ Commented Aug 20, 2021 at 19:48
  • $\begingroup$ @Evan The displacement n the work-energy theorem is not a generic displacement but the displacement along the actual trajectory. At each time, the position and the velocity depend on the initial conditions and force at an infinitesimal previous time. Once one has position and velocity, bot work and change of kinetic energy can be obtained at the same time. No one of them hames before the other. $\endgroup$ Commented Aug 20, 2021 at 19:56

Work done is energy transferred. So the energy transferred to the body when it is accelerated is the work done. Therefore the the work done and the kinetic energy increase at the same time.


Work is a transfer of energy by any means that does not depend on a temperature difference.

Therefore, when work occurs energy is transferred. There can be no first or second in this. They are always and must always be at the same time.

It is possible for work to immediately cause a change in potential energy instead of a change of kinetic energy. But whatever the form of the transferred energy, the change in energy occurs simultaneously with the work.

If that were not the case then energy would not be conserved.


One should remember how we define energy in physics, energy is defined as “ the capacity of an object to do work,” basically,the amount of work done on an object is equal to the gain/loss in the kinetic energy of the object.Say,as an example;I push an object,Now; how much work I did would be equal to the energy(in this case,kinetic energy) of the object.For further reading,I recommend reading about the work-energy theorem to which I am going to link a Wikipedia article below. Work-Energy Theorem(WikiArticle ): https://en.m.wikipedia.org/wiki/Work_(physics)


You have some good correct answers, but you seem to be concerned about the timing. Action and reaction of forces occur simultaneously. Changes in gravitational or electromagnetic forces can affect the object at the speed of light, contact forces will transmit through the object at its speed of sound, but as soon as each atom is accelerated its KE changes.


Earlier the body was at rest. When force was applied, it gains some velocity and thus gains kinetic energy and after gaining velocity, the body moves.

So doesn't that mean that the body gained kinetic energy first and then work was done ?

As soon as the body moves it has kinetic energy. The kinetic energy of the body is $\frac{mv^2}{2}$ where $v$ is the instantaneous velocity. As the velocity increases so does the kinetic energy possessed by the body.

But the body will not attain a velocity from rest unless energy is transferred to the body from something else to accelerate it. That something else is a net force acting through a distance per Newton's second law.

So work comes first.


My answer is with respect to real bodies, not ideal rigid bodies which do not exist at the macroscopic level. Real bodies don't acquire velocity (and thus kinetic energy) instantaneously (in zero time).

Hope this helps.

  • $\begingroup$ so what comes first ? $\endgroup$
    – Ankit
    Commented Jan 15, 2021 at 17:50
  • 1
    $\begingroup$ The body might already have a kinetic energy, while work be still zero. As the energy is transferred it results in change in KE or gain in KE, but it's the same as work. So how can we say work occurs first? They are both happning together. $\endgroup$
    – Proxy
    Commented Jan 15, 2021 at 18:11
  • 2
    $\begingroup$ It was addressed to you. The Work Energy Theorem is derived from Newton's second law. So saying that work and change in kinetic energy are not instantaneous is equivalent to saying that force and acceleration are not instantaneous (and let's keep the discussion to point particles). $\endgroup$ Commented Jan 15, 2021 at 18:47
  • 1
    $\begingroup$ @Bob D why did you mention that real bodies don't acquire velocity instantly ? $\endgroup$
    – Ankit
    Commented Jan 15, 2021 at 20:37
  • 1
    $\begingroup$ @BobD ...Those masses start moving as soon as the springs make contact, it's just that the initial force is so low (with such a unstiff spring) that the acceleration on the other block is very low, and therefore it barely moves when they make initial contact. Similarly with a mass deforming, it often does slightly shift the centre of mass. Parts of the non-rigid body at least have to gain kinetic energy for them to be deforming, and therefore kinetic energy is changing, since parts of the body with mass are clearly moving as a result of deformation. $\endgroup$
    – JMac
    Commented Jan 15, 2021 at 21:57

You've expressed an interesting dilemma: Work involves the application of a force through a distance. Also, considering the example of a block on a frictionless table, work done equals the change in its kinetic energy. So I think you're saying that work can't be done until the block moves through some distance but the block won't move until work is done on it.

But when the force is first applied, by Newton's second law there is an acceleration, i.e., a change in velocity, and the force and acceleration occur simultaneously. The force is what causes the block to start moving and we can calculate its subsequent motion and kinetic energy using Newton's law if we know the force.

Let's say the block moves from x=0 to x=$x_f$. The work/energy relationship is a way of calculating the kinetic energy and therefore the velocity of the block at x=$x_f$ without having to be concerned with the motion of the block at every point along the way. So to return to your dilemma, I think it arises by thinking it is the work done on the block that is causing it to move. Rather it is the force that is doing that.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.