# Work = Force x Distance vs Displacement

The difference in using Distance vs Displacement is demonstrated in this example:

Work = Force x Distance

If I carry an object to and fro 10 metres, the work done would be Force x 20 metres.

and

Work = Force x Displacement

If I carry an object to and fro 10 metres, the work done would be Force x 0 metres.

In this context, which should be a more accurate representation or formula? I note that Force and displacement are vectors and distance as scalar.

• The correct formula in the second case is $W=\int_{\vec{x}_0}^{\vec{x}_1}\vec{F}\cdot\vec{dx}$. This is not the same as $\vec{F}\cdot(\vec{x}_1-\vec{x}_0)$ since $\vec{F}$ is not constant (the direction of $F$ reverses after the first 10 m). May 18, 2015 at 13:18
• "(the direction of F reverses after the first 10 m" @lemon. That really depends on the nature of the force for which the work is being calculated, so you should not make blanket statements like that. May 18, 2015 at 15:42

It depends on whether the force field is conservative or not.

Example of a conservative force is gravity. Lifting, then lowering an object against gravity results in zero net work against gravity.

Friction is non-conservative: the force is always in the direction opposite to the motion. Moving 10 m one way, you do work. Moving back 10 m, you do more work.

As @lemon pointed out in a comment, this is expressed by writing the work done as the integral:

$$W = \int \vec F \cdot d\vec{x}$$

When $F$ is only a function of position and $\vec \nabla\times \vec F = 0$, this integral is independent of the path and depends only on the end points; but if it is a function of direction of motion, you can no longer do the integral without taking the path into account.

• What is the force vector in the gravity case? You are not pushing it down, you are resisting its downward movement. May 19, 2015 at 4:07
• @Random832 The force vector points down in the case of gravity - so when you are lifting, you are doing work against gravity; and when you are lowering it, gravity is doing work on you. While it is hard to understand that when you think about your muscles (which are not so good at storing and returning energy) it's easier to see that this works with a spring. A mass on a spring can bounce up and down against gravity; at the bottom of the motion, the spring is extended (contains stored energy); at the top of the motion the spring is less stretched, and the energy is gravitational P.E. May 19, 2015 at 4:25
• "When F is only a function of position, this integral is independent of the path and depends only on the end points" No, the curl of F field must be zero everywhere for this to be true. May 19, 2015 at 7:06
• @SiyuanRen - you are right, I have updated my answer to state this explicitly. Thank you. May 19, 2015 at 13:10

If you 'carry' an object 10 meters in one direction then return it back 10 meters from where you started the work done on the object is not the force you expended times distance walked.

The formula you write is often misunderstood and misused. In your example, when you lift the object in a gravitational field, the work being done on the object is its weight (force) times the vertical distance it's lifted. When you walk ten feet in one direction, then back there is zero work done on the object assuming you held the object at the same vertical position. Once you return to your point of origin and lower the object the object is now doing work on your muscles.

The process of carrying the object horizontally is much more complicated when trying to figure out how much work is being expended. The expended work is what's being spent in maintaining the isometric tension in your muscles. It's not force times distance.

But returning to your original question I believe you are having difficulty differentiating the terms distance and displacement. Distance is an absolute measure between two points in space. Displacement is a relative measure of distance which can turn out to be zero if you start and end at the same place.

When you raise a weight 10 ft high and lower it to it's original height the net work done ON the object is zero since the displacement is zero. That's because when you return the object to the original position its gained potential energy (that was stored 'in' the object) is being spent. The object is now doing work on whatever opposing force is doing the lowering.

The formula can be used when the restoring force is other than a gravitational field. A spring for example allows you to store then release energy when it's compressed and expanded respectively. Work is done on the spring, and the spring subsequently does work on the object.

Displacement refers to the object's position relative to the observer. The "place in space" of the orange.

Distance is the object's position relative to an earlier position.

If you pick up an orange, and run 10 miles holding it straight out, no work gets done on the orange. But if you extended your arm 10 miles, you would have to be doing work on the orange.