A Sample Picture of the Problem

Seem, A person is moving a object of Having mass m kg, Applying F Newton Force in A Circular path. The Radius of the Circle is r metre.

What is the work done if the person starts From A point of the circle and Comes back again on that Point after crossing whole the circumference of the Circle?

The person has applied the force for all the time.

I am confused about the displacement here.

What is the crossed Displacement here? Is it 0 metre or 2πr metre?

[N.B: It's a Circular path, The Object is not moving in Circular motion] I searched on google, But didn't get my answer. I am in grade 10.

I know that,


But I am confused with the displacement here! Kindly Explain me please. Thanks

  • $\begingroup$ buddy displacement is $0$, but to calculate work done we consider displacement along direction of force which is $2\pi r$ so work done is $F2\pi r$ $\endgroup$
    – maverick
    Commented Apr 10, 2020 at 6:21

1 Answer 1


For this answer I will assume that the person is pushing on the box with a force $F$ in the direction of motion (tangent to the circle).

The work done is given by $$ W = \int \vec F \cdot d\vec s$$ where $\vec F$ is the vector force applied and $d \vec s$ is the infinitesimal vector displacement. This is a path integral.

Since the direction of the force is changing, we can't just change the integral into $|F||s|$. The total displacement in this case, since you returned to the starting point, is exactly zero.

We have to consider the force and the displacement at each point along the path (with calculus). Fortunately, there is a simplification to avoid the calculus. Let's move to polar coordinates. $$\vec F = F_0 \hat \theta$$ $$d\vec s = r d\theta \, \hat \theta $$ So we can combine and integrate $\theta$ from $0\to2\pi$. $$W = \int \limits_0^{2\pi} F_0 r d\theta \, (\hat \theta \cdot \hat \theta) $$ The advantage here is that in polar coordinates, $\vec F$ and $d \vec s$ are constant. When the integrating a constant, we can simplify to just multiplication: $$W = F_0 r \theta |_0^{2\pi}= F_0 r \times 2 \pi$$ Notably, this is just $F$ times the total path length $2\pi r$.

Non-calc version: The work is the sum of $\vec F \cdot \Delta \vec x$ at each point on the path, if $\theta$ is the angle between the force and the displacement (direction of motion) and $\theta$ is constant, then the work done is $Fd\cos(\theta)$. Since here the force is always parallel to the direction of motion, its just $Fd$ where $d$ is the total path length ($d=2\pi r$).

Summary: The key point here is the dot product of the force and displacement at each point along the path. If you considered the normal force exerted by the ground (which is perpendicular to the displacement of the box), it would have done no work on the box.

  • 1
    $\begingroup$ Actually I study in 10th Grade and didn't study calculus as it's not included in our book. May you please just say in one sentence what should be the workdone? 0 or F×2πr? Sorry for my incapability $\endgroup$
    – Sindid
    Commented Apr 11, 2020 at 2:54
  • $\begingroup$ I added a noncalc version to the answer. $\endgroup$ Commented Apr 11, 2020 at 5:03
  • $\begingroup$ Ok I understand,but what will be the workdone if the person moves the object a partial way of the circle?Like from a Point 'A' to an another point 'B' of the circumference? $\endgroup$
    – Sindid
    Commented Apr 12, 2020 at 6:29
  • $\begingroup$ It's just whatever fraction of the distance traveled. If they move halfway around the circle is $\frac{1}{2} F\times 2\pi r$. Crucially, the distance is the distance along the path, $\pi r$, not the final displacement ($2r$). $\endgroup$ Commented Apr 12, 2020 at 7:36

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