# Work is force into distance or displacement

1. Oct 29, 2013

### emmfranklin

Respected all,

If i push an object on the floor and drag it say 2 m in a straight line with force y. then the work done is force times displacement.
that is 2y.

first doubt

which formula is correct one
work = force multiplied by displacement
or
work = force multiplied by distance.

since most websites say ... work = force multiplied by displacement

hence

in my given example both the displacement and distance is the same. so ill get the same answer.

but

now if i push my box in a circle track and bring it back to the starting point . the distance travelled might be say 20 meters
but the displacement will be 0 meters.

so what will be my work done.

according to the above formulas

it can be either
20 y joules

or

0 joules

which is correct .

is it work done or not

2. Oct 29, 2013

### arildno

For a force of constant magnitude, working constantly in the strict direction of the velocity of the object (however that changes), the work done by that force is magnitude of force times distance travelled.

I'm not sure where you've read anything else (I'm not doubting you!), but I will at least inform you that English Wikipedia, in the section Mathematical Calculation, has the correct relationship.

Last edited: Oct 29, 2013
3. Oct 29, 2013

### sophiecentaur

This should be clear cut - but I don't think it is. If you are considering what we call the Useful Work Done On an object then the Force times Displacement formula will tell you. This is fine when you are dealing with a conservative field - like when you are taking an ideal mass up to the top of an ideal hill or moving a charge from one object to another, through space.
But, where there is not a conservative field, the work done should be calculated by integrating along the path of the motion. The work done on a real car, getting it to the top of a real hill, will very much depend upon the route taken.
I would advise not getting too tied up with which of the two methods is 'correct' and consider the context of the actual problem before choosing which to apply. Most real world problems are concerned more with Work Done By but Work Done On will give a good idea of the energy you might get back from the object you just lifted up.

4. Oct 29, 2013

### arildno

sophiecentaur is giving the proper formulation for the GENERAL case.
What I wrote is valid for the SPECIAL case that I outlined:
The work done by a force that a) Has CONSTANT magnitude and b) Remains strictly tangential to the particle's path, that work equals (magnitude of) force times distance.

In integral form:
$$W=\int_{t_{0}}^{t_{1}}\vec{F}\cdot\vec{v}dt=\int_{t_{0}}^{t_{1}} F\frac{\vec{v}}{||\vec{v}||}\cdot\vec{v}dt=\int_{t_{0}}^{t_{1}}F||\vec{v}||dt=F\int_{t_{0}}^{t_{1}}||\vec{v}||dt$$

The integral of the SPEED over the time interval equals the distance travelled.
(If the force is always antiparallell to the velocity (say, as in kinetic friction), but constant in magnitude, add a minus sign in the expression.)

Last edited: Oct 29, 2013
5. Oct 29, 2013

### sophiecentaur

I think the OP was asking why there seem to be two answers to the same simple question. I was (we are) just pointing out that the answer depends upon what you actually want to know in each case and that you should not slavishly follow one or the other approach.
Many students write in, asking for a general rule about lots of things. It is very risky to try to apply the same rule in every case. Half the time, you could be wrong! That doesn't mean basic Physics is a matter of choice - it means that you often need to dig deeper than the surface to find how to apply it.

6. Oct 29, 2013

### arildno

Sure!
That's why I felt it important in the second post to bring about explictly that MY case is a special case of what YOU wrote, by deriving my result from the general definition of work.

7. Oct 29, 2013