Is the work-energy theorem valid when there's an impulsive force during motion of considered body?

For eg: Consider a man jumping from some height into a swimming pool of certain depth, if we apply work energy theorem from his initial position to final position (Change in Kinetic Energy would be 0), Should we consider the impulsive force that water would provide when man hits the surface of water?

Even if for an instant this impulsive force should act in opposite direction of man's motion there is some displacement and once he enters water buoyancy force would take over, Am I Getting It Correct?


3 Answers 3


Work requires both force and displacement. You could say that the displacement is a result of the force, which depends on the duration that the force is applied.

So work done might be small, even though a huge force has acted. If that huge force just applied at an instant, as an impulse, then the work is not as big as we would expect. But some work is still done.

How much this work, however small or big, influences a free object, is described by the work-energy theorem. This theorem combines the work $W_{total}$ done by external forces $\vec F$ on an object (the energy added to it) with the object itself (it's mass $m$) and it's mechanical state (it's speed $v$ and position $\vec s$). And that combination is done through energy conservation:

$$\begin{align} W_{total}&=\Delta K\\ \int \vec F \cdot d\vec s&=\frac12 mv_2^2-\frac12 mv_1^2 \end{align}$$

Work is in general an integral. Meaning, it is a sum of all force-times-displacement products at every instant, however small. So at impact with the water surface in your example, the large force that acts for a very short time (it acts over a tiny displacement) is indeed included and added to the total amount of work. While the object sinks further, the fluid resistance, buoyancy force etc. add additional work.

Which that is more significant depends on many things, but in the end they must all be added together into the total work done that causes the change in kinetic energy as the theorem tells.


If you're asking "Why can I deliver an impulse force (one that changes momentum in an infinitesimal time, implying there is no displacement) that does no work yet still changes the momentum of a body?"

Then you are right to be concerned about the validation of the theorem, but the problem is not in the theorem but in modeling a force that occurs over no time that still imparts momentum. If we look at the equation for a change in momentum,

$$J=\int^{t_2}_{t_1}F~dt $$

If $t_1=t_2=$ then $J$ would be zero (unless the force were infinite over that point in time). But if we just say that the force impacts over a very short time, $t_2-t_1<<1$, then we can have a nonzero change in momentum, and the body will be displaced as the momentum is delivered.

The problem comes in where you have to know the force as a function of time or as a function of position and because these are small values they usually aren't easy to gather experimentally with enough accuracy and if this is a physics problem and the curve can be supplied, it might require integration for a very small change in the answer.

At the end of the day, we can make a modeling assumption that momentum will be conserved over an infinitesimal distance and time to calculate momentum after (or before or by the impact depending on how it's asked), from that behavior we can back out the value for the energy delivered by the impulse. To make the problem even simpler, if the average force is given and the duration of impact is given, we will just assume the body does not displace in this time and we can compute the change in momentum, and energy delivered algebraically.

  1. You are totally correct about your reasoning. Remember, the work-energy conservation theorem is always valid.

  2. Thus, whenever you see a change in the total energy of a system, it means the energy goes into/out of that system. But you can always include more "objects" into your system and observe that the conservation of energy still holds.

  3. Back to your question: The moment the man hits the water, his kinetic energy decreases because his velocity decreases. The escaped kinetic energy transfers into internal and kinetic energy of the water in the swimming pool.

  4. When buoyancy force takes over, the potential energy of the man-swimming pool system increases because the man is being pushed upward. But some water molecules are pushed downward at the same time, and also the pressure the body of water exerts on the pool walls increases. Of course, there are many other processes happening at that moment, but remember the conservation of energy is always true.

  5. When the man's velocity decreases to zero, note that the height of the water body in the swimming pool increases by a bit. Also, the friction force that took place between the man and the water increases the internal energy of the system, making the water "hotter". Again, there are many other processes that has happened during that period of time but the conservation of energy must always holds true.

  • $\begingroup$ If the man starts at $0 \frac m s$ and ends at $0 \frac m s$ the net change in kinetic energy of the man over that time frame is definitely $0$. $\endgroup$
    – JMac
    Oct 19, 2017 at 15:14
  • $\begingroup$ @JMac That's why I said "before he has hit the water". $\endgroup$ Oct 19, 2017 at 15:16
  • $\begingroup$ But his statement was "if we apply work energy theorem from his initial position to final position (Change in Kinetic Energy would be 0)". You said "this is incorrect"; but it is not. You took the line completely out of context and then said it made him wrong... $\endgroup$
    – JMac
    Oct 19, 2017 at 15:17
  • $\begingroup$ @You are correct... I should have read more carefully. See edit. $\endgroup$ Oct 19, 2017 at 15:30

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