Isn't the string pulling the ball upwards, and, therefore, doing work on the ball, like when someone lifts weights. I understand that the force is perpendicular but can I get an intuitive answer as to why it pulls the ball up, yet is doing zero work?

  • 2
    $\begingroup$ Exactly for the same reason why Lorentz force does no work: physics.stackexchange.com/q/274459/226902 physics.stackexchange.com/a/566528/226902 $\endgroup$
    – Quillo
    Mar 7 at 19:06
  • 3
    $\begingroup$ Ah, the sweet sweet smell of homework. What happens to the total energy of the ball as it moves? $\endgroup$
    – Boba Fit
    Mar 7 at 23:03
  • 1
    $\begingroup$ The OG of caveats : "Newton’s first law states that every object will remain at rest or in uniform motion in a straight line unless compelled to change its state by the action of an external force." ... force : "the quantities causing an acceleration of a body". ... work : "The product of the force on an object and the displacement the object undergoes along the direction of the force." - A pebble in a river is under considerable forces, but until it washes away, no work is being done to it or by it. $\endgroup$
    – Mazura
    Mar 8 at 1:45

10 Answers 10


Your intuition seems to conflate work with force. But just because a force is present, that doesn't necessarily mean that it does any work.

Just like when you push hard on a wall - great force but no work was done (nothing was changed by your efforts).

Work requires two components to be present: force and displacement. The formula (in case of constant force) is $$W=\mathbf F\cdot \mathbf r\,.$$ Think of pushing on a train cart rolling on tracks:

  • When you push along with the tracks, then your force causes a displacement of the cart (it moves). You (your force) have now done work on the cart (added energy to the cart, in this case converted to kinetic/motion energy).

  • But if you push sideways to the tracks, then the cart isn't moving and no displacement happens. So no work is done. Even if any displacement is taking place while you are pushing, then it certainly is not a result of your force. Because your force is perpendicular to this displacement. Whatever energy you may have spent on producing your force is just lost (as heat or work to the surroundings or internally within your body) - nothing was transfered as work to the cart.

In the same way, the string in your scenario pulls always perpendicular in the pendulum and so never pulls along with the displacement direction and thus does not work.

The concept of work can often be confusing to get the intuitive sense of. But think of it as mechanical energy transfer which requires a position change (in contrast to heat which is thermodynamic energy transfer and in general requires a temperature change). Then it might be easier to intuitively grasp.

  • 2
    $\begingroup$ When someone presses hard on a wall, their muscles will expend energy and they will eventually become tired. Is this not work? Similarly with holding one's arms horizontally - at right-angles to the body.. $\endgroup$ Mar 7 at 9:20
  • 5
    $\begingroup$ Yes because the muscle is bag of small proteins that interact weakly and need to be held together by constantly running chemical reactions. A string is held together by stronger chemical bonds so no chemical energy is spent by just holding. The two links in my answer below might be interesting for you $\endgroup$
    – akraf
    Mar 7 at 9:32
  • 13
    $\begingroup$ @chasly-supportsMonica Sure, this is work being done. But it is being done within the body, not on the wall. And the goal was to do work on the wall. In the answer I mention that such energy spent to produce the force is just wasted if it does no work on the object. The human body is indeed a special case that spends energy in order to produce a pushing force. In contrast, no energy is spent when a ladder is leaning up against and thus pushing on the wall. The fact that energy is spent in one case but not the other, is not about the applied force, but about how this force is produced. $\endgroup$
    – Steeven
    Mar 7 at 9:56
  • 6
    $\begingroup$ @chasly-supportsMonica Feynman, in sect 14-1 of his physics lectures call this "physiological work" to distinguish it from physics work which, in the case of pushing on a wall or simply holding on to a (and not moving) weight is zero. The energy expended is all internal to the body. $\endgroup$
    – Bob D
    Mar 7 at 13:36
  • 4
    $\begingroup$ comments like chasly's (which is spot-on!) are why using people pushing/holding/carrying things make bad examples when discussing work. Much more intuitive are objects on a table, or in this case, hanging from a rope. It's obvious there's not energy expenditure for an object hanging from a rope at rest $\endgroup$ Mar 7 at 23:30

The mathematical reason:

The work done by tension is zero because the force by the string and the displacement on the ball are perpendicular to each other.

The intuitive reason:

Any change in the speed of the ball is caused by gravity. The ball goes upwards with a decreasing speed because it gains gravitational potential energy at the cost of its kinetic energy. The string is a one-dimensional object and hence can't push or pull the ball along its curved path. All the string does is pull the ball towards the center of the curved path, thereby changing the direction of motion of the ball, not its speed. Since the speed (and therefore kinetic energy) of the ball is unaffected by the string, no work is done by the string's tension force.


Work is force times displacement in the direction of the force or $W=Fd\cos\theta$ where $\theta$ is the angle between the force and displacement. The pendulum tension is a centripetal force which acts perpendicular (at right angles) to the displacement. Thus $\cos \theta =0$, and $W=0$.

Hope this helps.


@Steven already notes in his answer that you may think intuitively that holding a force/tension needs work.

That might come from the experience that excerting a force with your body spends energy, e.g. when holding up a weight. And indeed chemical energy is spent in this special case, because the filaments of the muscle cells are not one long band but consist of many proteins that basically need to be held together by continuously running chemical reactions. But in your case, there is no muscle, but a string.

That is further explained for example in this Stack Overflow question: Why does holding something up cost energy while no work is being done?

A visual idea of how muscles have to spend energy to maintain tension might be found in this video about the cross-bridge cycle: https://youtu.be/99R-XCGme8Q?t=62

But in your example the string holds the ball, not the body


In a very intuitive way, doing work requires some energy source. Work is the transfer of energy, you can't perform work without the energy coming from somewhere. That energy source could be in the form of chemical bonds like in gasoline that powers an engine or in food that powers your body, or in a physical form like elastic potential stored in a spring or gravitational potential stored in a weight placed on a tall shelf, or in other forms. You have to ask, if the string is in fact doing work, where does that energy come from?

The problem is, a massless, inelastic string has no power source, and no way of storing energy. Therefore, it is impossible for the string to ever do any work at all, no matter what you do with it. For the string to do work on the pendulum bob, energy would have to move from the string to the bob, but the string cannot store any energy in the first place. Even if you were to tie the string to a weight and pull directly upwards, it would be you performing work on string + weight system, but not really the string performing work on the weight.


The string doing no work at any moment. Lets pick one moment to think about - say when the ball is moving away from the centre to the right and about half way up.

At that point the string is pulling the ball up and to the left - so we can think of it as doing two things at once - pulling the ball left, so slowing the its rightward motion, and pulling it upwards. Pulling it upwards would be doing positive work, adding energy, and pulling left, slowing its motion is doing negative work, removing energy.

Add these two together and you find that the string is doing doing zero total work at any moment. Other answers go into the maths, but whenever we have a diagonal force it can be easier to analyse if we think of it as a combination of an up/down force and a right/left force.

And of course the same argument applies for every moment as the pendulum swings.


The length of the string does not change, so the tension in the string does no work.

Work is force times displacement of the point of application of the force in the direction of the force (or directly opposite to it, which can represented with a minus sign.)

It's like when an object orbits the earth in a circle. The distance to the center of the earth doesn't change, and so no work is done. And the force is always perpendicular the direction that the object is moving. Note that it has to be a circular orbit.

It's not like someone lifting a weight in a gym, rather it is like someone holding a weight with his arm hanging straight down and then swinging it back and forth slightly. Effortless. Or at least no harder than supporting it in a stationary position.

Or consider a mass moving in a circle, e.g. a bob on the end of a massless, fixed-length string that is attached to a point that cannot move. The string exerts a force on the bob that makes it move in a circle. It does not do any work because the bob does not change speed, only direction. This is related to the fact that the force on the bob is always perpendicular to the direction of the bob's motion. The string constrains the bob to move along a particular path, but it doesn't add or remove any energy from the bob.

Going back to the pendulum, imagine that instead of a string constraining the bob to move to and fro along a part of a circle, you have half a hemisphere made of rigid material that is fixed immovably to the earth. The bob slides about frictionlessly inside the bowl, moving to and fro along a part of a circle, exactly as it did with the pendulum. Let's consider the bob to be a point mass, to keep things simple.

It is intuitively obvious, I hope, that the bowl does not do any work on the bob. The bowl is rigid and immovable. The total energy of the bob is constant. All that changes is the fraction of the bob's energy that is kinetic energy, and the fraction that is gravitational potential energy.

Just as a mass in free fall has kinetic energy and potential energy, and the gravitational field only transforms one form of energy to the other, with the total being constant, so it is with the pendulum and with the point mass sliding in the bowl frictionlessly. There is work being done, but only by the gravitational field.

Consider a point mass sliding up an inclined plain, slowing down as it does so, due to the gravitational field. If there is a sense in which your pendulum string is pulling the bob upwards, in the same way, the inclined plane is pushing the mass up. This is especially clear if you imagine an inclined plane that is at very small angle away from horizontal. It should be intuitive now that inclined plane is passive, unmoving, rigid, and indeed immovable, and is not doing any work. If anything, the bob is doing work on the gravitational field. But since this is equivalent to storing potential energy in the bob, the total energy of the bob remains the same.


Let me give you an example for a field which exerts a force on the particle but doesn't do any work. Consider the magnetic Lorentz force $$\vec{F} = q(\vec{v} \times \vec{B})$$. where $\vec{B}$ is the external magnetic field which is present in the region. Now consider the work done by this force. $$W = \int \vec{F} \cdot \vec{dl}$$. This could be further be written as $$W = \int q(\vec{v} \times \vec{B}) \vec{dl}$$. We can also write $\vec{dl} = \vec{v} dt$ which after substitution would be $$W = \int q(\vec{v} \times \vec{B}) \cdot \vec{v}dt$$ We can immediately notice that the term $$(\vec{v} \times \vec{B}) \cdot (\vec{v} dt) = 0$$

This is because $\vec{v} \times \vec{B}$ is always perpendicular to $\vec{v}$. Therefore $$W = 0$$ for magnetic fields.

Therefore, you can see that although there is force on the charged particle due to the magnetic field, it doesn't necessarily mean that the the field is doing work on the particle. It's the $$\vec{F} \cdot d\vec{l}$$ that matters when we are concerned about the work done by something.


You already have some good answers, but I would like to take a different angle.

The work done by the tension in an ideal pendulum string is zero because of the standard assumptions, including that the "string" is really a massless, non-elastic rod.

If you would like the string to do some work, change it to a non-ideal material. For example, use a real rod or string with some non-ideal elasticity that can convert longitudinal force to heat by internal friction. When the force along the string reaches its maximum (the low point of each swing), the rod reaches maximum extension. Since the elasticity is not ideal, the rod doesn't return all the energy to the motion of the bob on the upswing. It's doing work at each oscillation.

Because the tension in a real pendulum made of real materials does non-zero work, it loses energy and the pendulum motion eventually decays.

  • 1
    $\begingroup$ Even if the string is lossless but elastic there would still be energy being transferred into and out of the string periodically. $\endgroup$
    – TimWescott
    Mar 7 at 18:59
  • 1
    $\begingroup$ @TimWescott True, but passing work back and forth between two ideal objects ad infinitum is no fun, and not in the spirit of my answer. $\endgroup$
    – Theodore
    Mar 10 at 14:07
  • $\begingroup$ Granted, but I felt that someone coming to the subject cold might read your answer and think that any string with "give" would lead to dissipation. $\endgroup$
    – TimWescott
    Mar 10 at 16:41

Only if the target force has a component force in the direction of movement, the force does work to the object.

Here no component of tension acts in the direction of movement of pendulum, so the work done by tension is $0$.

Simple pendulum


Your Answer

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

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