Consider an object being pushed 3/4 of the distance around a circular track. The work done on the object would be the distance of 3/4 the track’s circumference times the force applied to the object (given that it was pushed at a constant force). Since we are multiplying a vector by a scalar, why is work a scalar measurement? Or would the work done on the object actually just be force times displacement? Thanks.


4 Answers 4


Work is the dot product of a vector force and a vector displacement, hence a scalar.

Knowing just the scalar distance isn’t enough to calculate work. That distance might be in the same direction as the force, but it might be perpendicular or even opposed. All of those would give different values for the work done.


The general definition of work is $$W=\int\mathbf F\cdot\text d\mathbf x$$ Which essentially says, "Add up all of the dot products between the vector force $\mathbf F$ and the vector displacement $\text d\mathbf x$ along the path the object travels on." Since we are adding up dot products, which are scalar quantities, the work done by a force is also a scalar quantity.

The confusion might arise with specific cases. For example, if the force is always points parallel to the path, then the dot product becomes the product of the magnitudes $$W=\int F\,\text dx$$

And then if the force is constant in magnitude we get the "algebra based physics work" $$W=F\int\text dx=F\Delta x$$

But now we have a scalar force magnitude $F$ multiplied by a scalar distance $\Delta x$. In this equation $F$ is not a vector, but rather its magnitude, and $\Delta x$ is not a vector, but rather it is the total path length.

With work you will never have a vector multiplied by a scalar, since that will result in a vector quantity, which doesn't work (pun always intended) because work is a scalar quantity.

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    $\begingroup$ Thanks for the detailed answer! So work would really be the dot product of the force and displacement? Would this mean that if, say, an object were pushed around a lap and ended where it started, the total work applied to the object would be 0 joules? It seems counterintuitive, but I see how it works. $\endgroup$
    – R_4127
    Mar 17, 2020 at 4:58
  • $\begingroup$ @R_4127 No, it would not be $0$. Given how we got to the final equation $\Delta x$ is not the displacement from the start to the end, it is the distance traveled along the path, which is not $0$. $\endgroup$ Mar 17, 2020 at 12:24
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    $\begingroup$ @R_4127 It is the length of the path. It doesn't have anything to do with a displacement (as in $x_\text{final}-x_\text{initial}$). This is why I went through the "derivation" of the equation in my answer. Before you use a physics formula, you have to make sure you know what the variables mean and when the equation can be applied. Hopefully my answer helps give that context :) $\endgroup$ Mar 17, 2020 at 16:14
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    $\begingroup$ @Graham Actually in my final equation $\Delta x$ is the path length. $\endgroup$ Mar 17, 2020 at 21:59
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    $\begingroup$ @DavidReidy That is completely incorrect. Please do not mislead the OP. Just because an object starts and stops at the same position does not mean the work done by a certain force is $0$. Also in my final equation $\Delta x$ is not the displacement between the start an end of the path. And finally, physics definitely can take into account friction and air resistance. $\endgroup$ Mar 18, 2020 at 12:28

The work is the dot product of the force and the displacement, and displacement is a vector; we have to take into account what direction it is pointing. If an object is traveling in a circle, then it has to have a centripetal force, so it doesn't have a constant force. It could have a constant magnitude force, though. The centripetal force is perpendicular to the displacement, so the work done by the centripetal force is zero.

If you look at a ball thrown at an angle of 45 degrees, the ball's velocity starts out with an upwards component, while gravity is pointing down. Since the angle between them is more than 90 degrees, he dot product is negative; gravity is decreasing the kinetic energy of the ball. At the peak of its trajectory, the velocity is perpendicular to gravity, and so, at the very moment, gravity is not doing any work (you can verify that by writing an equation for its kinetic energy in terms of time, then taking the derivative with respect to time). Once the balls starts coming back down, the angle between its velocity and gravity is less than 90 degrees, so gravity is doing work on it, and it is speeding up.

  • $\begingroup$ "If an object is traveling in a circle, then it has to have a centripetal force, so it doesn't have a constant force." It doesn't have a constant net force acting on it. But a constant force can still be acting on it. This answer slightly is misleading because it suggests that the only work that you consider is the net work, which is not the only case. Circular motion, even if it is uniform circular motion, does not necessarily mean any work you look at will be $0$. $\endgroup$ Mar 19, 2020 at 21:52
  • $\begingroup$ @AaronStevens "It doesn't have a constant net force acting on it." That's what "force", without qualification, means. Like, if I say "My weight is X", I don't need to specify "The weight of my entire body". "Circular motion, even if it is uniform circular motion, does not necessarily mean any work you look at will be 0." That's why I specifically said "the work done by the centripetal force is zero." $\endgroup$ Mar 19, 2020 at 23:02

The reason for this is that your understanding of the definition of work contains an error:

"product of ... distance (scalar)"

Work is not defined using a product of force and a "distance", but a displacement. A displacement is the difference of two positions, which are points, and displacements are vectors. So it is actually the dot product of two vectors, which, in turn, is a scalar.

That said, I think what you may be asking about here is that there is sometimes seen a simple scalar-only formula that looks like

$$W = Fd$$

where we involve only the (scalar) magnitudes of both force and displacement. This formula only works in one dimension(*), or else that the force and displacement are acting along the same line. Otherwise, it looks like

$$W = Fd \cos \theta$$

where $\theta$ is now the angle between their lines of action, and this now is exactly the dot product of vectors.

There is no "vector times scalar" formula for work. Such a thing would, indeed, as you suggest, be a vector, and work is not a vector.

(*) Technically, in one dimension it is also a dot product of two vectors, but there is little functional difference between a vector of dimension 1 and a scalar, mathematically. That said, I still think it may be useful to keep this distinction in mind as a matter of conceptual clarity. A vector of dimension 1 has a solitary component:

$$\mathbf{v} = \langle v_x \rangle$$

but it belongs to a different "data type", so to speak, than scalars (real numbers) do and this has some important algebraic consequences such as that you cannot "sensibly" add a one-dimensional vectorial quantity and a scalar real number.


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