# Work done on a frictionless surface

Imagine that we apply a force $$F$$ on a frictionless surface to move a body by a distance $$d$$. (The body starts at rest and is stopped after moving a distance $$d$$.)

1. Is the work done $$F d$$? But from what I understand since the initial and final velocity of body is zero, the change in KE is also zero and hence work done on the body is zero. Positive work to move it and when it attains some speed, we have to apply a force to stop it and so negative work to stop the body.

2. If the same body is moved such that it is brought back to original position, then is the work done zero? Why? because it is a closed path or because the change in KE is zero?

We generally do work against some force (gravitational, frictional). Here none of these forces are there, so against what force are we doing work?

• How does the object stop after distance $d$? The surface is frictionless. Apr 19 at 7:14

You specify only a single force here that doesn't change at the beginning (you hint at a second force later), but in order to start and end at rest there would need to be some other force present to slow the object back down. So in order to have a valid system first let's say we have your applied constant force $$F$$ that turns off when the object stops and then some other force $$F'$$ that is in the opposite direction and is responsible for bringing the object to rest (for example, $$F'$$ could be larger than $$F$$ in magnitude but turn on at a later time).

Is the work done F x d ? But from what I understand since the initial and final velocity of body is zero, the change in KE is also zero and hence work done on the body is zero.

The net work would be zero ($$W_\text{net}=\Delta K$$), but this does not mean no forces do any work. The work done by $$F$$ and $$F'$$ will both be nonzero add up to be zero over the entire trajectory of the object here.

If the same body is moved such that it is brought back to original position, then is the work done zero? Why? because it is a closed path or because the change in KE is zero?

The net work is zero just like before since the change in kinetic energy is zero. But just because the path is closed does not mean the work done by a force has to be zero. Remember, $$W=F\times d$$ is only valid when you have a constant force in the direction of the displacement. For an object moving in a circle this cannot be the case for at least one of the forces acting on it. You would have to bring in calculus for the more general expression $$W=\int\mathbf F\cdot \text d\mathbf x$$, which for those who are less familiar with calculus is just saying "ok, so then let's break the path up into really tiny pieces so that on each piece the force is essentially constant, and then add up the work done along all of those little pieces".

We generally do work against some force (gravitational, frictional). Here none of these forces are there, so against what force are we doing work?

Work does not need to be done against any force, and honestly the phrase "work done against insert force here" is horrible terminology in my opinion. Forces do work. That is it. If you have two forces in opposition acting on a moving body each force will do their own work. It just adds confusion to say, for example, "work done against friction", since really what you mean is "work done by the applied force".

Imagine that we apply a force $$F$$ on a frictionless surface to move a body by a distance $$d$$. (The body starts at rest and is stopped after moving a distance $$d$$.)

If the force $$F$$ we apply is constant over the entire distance $$d$$, then the only way the body can stop at the end of the distance $$d$$ is for another greater force to be applied opposite to your force for some time (and distance) prior to the body reaching $$d$$, causing the body to decelerate and come to a stop at $$d$$.

1. Is the work done $$F d$$?

Yes, the work done by you in applying the force $$F$$ is $$Fd$$

But from what I understand since the initial and final velocity of body is zero, the change in KE is also zero and hence work done on the body is zero.

Yes, but it is the net work done on the body that equals zero, not the work you did on the body. The net work equals the positive work of $$Fd$$ that you did, plus the negative work done by the opposing force that resulted in bringing the body to rest at $$d$$. That negative work is $$-F_{ave}d$$ where $$F_{ave}$$ in this case is the average force exerted by the opposing force over the entire distance $$d$$ with $$F_{ave}=F$$. It is the average opposing force over the distance $$d$$ because it had to be greater than $$F$$ while it was applied to decelerate the body.

Positive work to move it and when it attains some speed, we have to apply a force to stop it and so negative work to stop the body.

Correct. The opposing force, while it was applied during, had to be greater than $$F$$ to decelerate the body doing negative work.

1. If the same body is moved such that it is brought back to original position, then is the work done zero? Why? because it is a closed path or because the change in KE is zero?

The net work is zero because the change in KE is zero. It doesn't have to be a "closed path". All that matters is that it starts at rest and ends at rest, or that its initial and final velocities are the same.

We generally do work against some force (gravitational, frictional). Here none of these forces are there, so against what force are we doing work?

To say "we generally do work against some force" is not true. You don't necessarily do work against forces that do negative work as in your example. If there was no opposing force to bring the body to a stop, then you would still be doing work of $$Fd$$. The difference is the body would have kinetic energy of $$\frac{1}{2}mv^{2}=Fd$$ after moving the distance $$d$$.

Hope this helps.