*I suggest that you read* post 16.13, post 16.16, post 17 *and* post 16.19 *before you read this one*.

A force does *work* when it moves something. If a constant force pushes or pulls something in the direction in which it pushes or pulls, the work that it does is defined to be the value of the force multiplied by the distance moved. So if a constant force of 1.5 N pulls something a distance of 2.0 m, the work done is 1.5 × 2.0 = 3.0 N.m. We expect to measure work in N.m because it is a force (measured in N) multiplied by a distance (measured in m). However, the units N.m have a special name when they are used to measure work – the joule (abbreviated to J). So we say that the work done by the force is 3.0 J.

Let’s think about pulling an object along a horizontal surface. If the object does not accelerate, the pulling force has only to overcome friction (post 16.19). This force is given by *F* = *mgμ _{k}*. In this equation

*m*is the mass of the object and

*g*is the acceleration due to gravity (post 16.16). So

*mg*is just the weight of the object (post 16.17). Experiments show that the force required to keep an object moving across a horizontal surface, at a constant velocity, depends on the weight of the object and the nature of the surfaces (the bottom of the object and the top of the horizontal surface) in contact. For a reasonably smooth block of copper moving on a steel surface, the force would typically be about 0.4 times the weight of the copper block; there is no theory that allows us to calculate this value – it’s simply an experimental observation. Then we call 0.4 the

*coefficient of kinetic friction*or

*coefficient of sliding friction*between the copper and steel surfaces. This coefficient of kinetic friction is represented by

*μ*in the equation in the second line of this paragraph. Since

_{k}*F*and

*mg*are both forces, they are both measured in newtons and

*μ*is just a number with no units; then the equation

_{k}*F*=

*mgμ*makes sense because it says that one force is equal to another force.

_{k}So how much work is done in pulling the object? If it moves a distance *L*, the work done is given by *W* = *F* × *L* = *mgμ _{k}L.* Suppose an object whose mass is 0.5 kg is pulled a distance 2.0 m along a horizontal surface at a constant speed. If the coefficient of kinetic friction between the object and the surface is 0.4, the work done is 0.5 × 9.8 × 0.4 × 2 = 40 J (remembering that the acceleration due to gravity is 9.8 m.s

^{-2}, see post 16.16). (Note that the answer to this simple calculation is not 39.2 J, as explained in post 16.7).

Now let’s think about lifting an object of mass *m* to a height *h*. To do this we have to overcome the force of gravity, *mg* (post 16.16), that is pulling it downwards. So the work done is given by *W _{g}* =

*mgh*which represents the force multiplied by the distance moved. Why have I used the symbol

*W*, to represent the work done, and not just

_{g}*W*? The reason is that I’ve used

*W*, in the previous paragraph, to represent the work done in pulling an object across a horizontal surface. This post wouldn’t make much sense if I used the same symbol to represent two different things! I could have chosen any symbol that I like (see post 16.13). I like

*W*because the

_{g}*W*helps remind us that it represents work and the subscript

*g*reminds us that it’s associated with gravity.

So let’s suppose that I lift a box whose mass is 1.5 kg to a height of 0.5 m. The work I do is 1.5 × 9.8 × 0.5 = 8 J.

The previous paragraph shows that someone who lifts boxes all day does much more work than any scientist that I’ve ever met!

*Related posts*

16.17 Weight

16.16 Gravity

16.19 Why do things stop moving….

16.13 Changes in movement

*Follow-up posts*