I read in many sites that the concept of mechanical energy is the ability of an object to do work, but how can an object do work? Isn't it rather the force applied to that object the one that produces work and not the object itself?. We can have an object that has mass $m$ and a constant speed $v$, with which, we can have kinetic energy $E_K = (1/2)mv^2$, but by work definition $W=\vec{F}d$ there's no work because there's no force applied. How is this explained mathematically?


3 Answers 3


Energy is the ability or potential to do work. In the case of an object with kinetic energy it can do work by exerting a force on another object (but it does not have to do work). This force could be exerted by colliding with another object, or by gravitational attraction, or it could be an electromagnetic force etc.

No matter what the force is, the work done on the second object is $W=\int \vec F . \vec {d s}$, which simplifies to $W=\vec F . \vec s$ if $\vec F$ is constant. By Newton’s third law the second object exerts a force $-\vec F$ on the first object and so does work $-W$ on the first object. So the first object loses an amount $W$ of energy.


TL;DR The kinetic energy is an abstraction of object's ability (potential) to do work. The object does not have to do work, but it could! As you have already indicated, if object keeps moving at constant velocity it does no work. But if object's velocity changes, then there is some work involved:

  • Object's velocity increases when there is non-zero acceleration in the same direction as velocity. This acceleration exists only if there is some force in the direction of motion, hence the work done by the (net) force is positive. It is said that work is done on the object as its kinetic energy increases.

  • Object's velocity decreases when there is non-zero acceleration in the direction opposite to velocity. This acceleration exists only if there is some force in direction opposite to motion, hence the work done by the (net) force is negative. It is said that the object does work as its kinetic energy decreases.

The work-energy theorem says that change in kinetic energy equals total work done on an object

$$\boxed{\Delta K = K_2 - K_1 = W}$$

where $W$ is work done by all forces, and it could be either positive or negative. If object's velocity decreases, then the energy goes away from the object to some other form of energy. If it all goes to gravitational potential or elastic potential energy, than it could be recuperated by simple means back to kinetic energy, which is exactly what the Law of conservation of mechanical energy says. If kinetic energy is lost to friction force, then the energy gets converted to heat and internal energy which cannot be, by simple means, recuperated back to kinetic energy. But if the object velocity does not change, the work is zero!

The same argument applies to gravitational potential and elastic potential energy. Emphasis on potential! Gravitational force does work only when the object moves vertically. But we define there is some energy in the earth-object system at certain height (altitude) even when there is no displacement in the vertical direction. It does not mean the gravitational force actually does work, but it could potentially do work if we allowed it to.


You can answer your question by doing as follows.

At first understand the significance and drawback of dimensional analysis/formula. More than two physical quantities can have same dimensional formula. One of the example already lies in your question. Others are:

  1. Work and Torque [ML2 T-2]
  2. Stress and pressure [ML-1 T-2]
  3. Surface tension and Surface energy [ML0 T-2]
  4. Velocity gradient and frequency [M0L0T-1]
  5. Refractive index and strain [M0L0T0]

So, now can you interchange things as of ex 4, absolutely not. If done so, the whole context changes.

Now understand the context that makes work and kinetic energy have same dimensions. Its because of integrating Newton's second law with respect to distance rather time. When this happens you would end up getting Work-Energy theorem.

So, if you want to understand work done mathematically which includes a force, then you express in integrals with respect to distance, for which the force displaces the body. In a few occasions you can express them in terms of change in Kinetic Energy( for any kind of force), negative of Potential Energy(for conservative force). Refer the book Mechanics by Keppler and Kolenkow for more help.

Formula Refrences

  • $\begingroup$ While torque and work/energy are different things, even though they have the same dimensions, work and energy are not: Work measures an amount of energy transferred. It's like the reading in gallons or liters you have at the gas pump as opposed to the reading on your fuel gauge: Both are measuring the same physical entity; it's just that the pump measures how much of it has been transferred. In all reality I'm not a fan of the work concept and haven't really used it past middle school -- it's like inventing a new name for liters flowing instead of sitting somewhere. $\endgroup$ Jan 27 at 10:32
  • $\begingroup$ As a formal remark: This site supports Latex in answers and comments, so you can write something like $Work\:energy = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta K \cdot E$, which renders as $Work\:energy = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2 = \Delta K \cdot E$. $\endgroup$ Jan 27 at 10:34
  • $\begingroup$ sorry, I have no idea what is Latex. Thank you for informing, I will check it out $\endgroup$ Jan 28 at 3:16
  • $\begingroup$ Just start like one always does: By copying/pasting, experimenting, googling ;-). Remember, physics.stackexchange gives you an instant preview of you answer while you are editing, so you can simply play around pretending that you are editing, and then cancel the edit. $\endgroup$ Jan 28 at 7:19

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