calculus-based physics for scientists and engineers

Work in One Dimension

An introduction to Work

Energy is hard to define right in the beginning. So, let's not. Your intuition about energy is pretty good though. Things moving fast have high energy; slow things have low energy. Batteries contain energy ready to unleash to power the world, but they lose energy as they do so. The sun bombards us with radiation energy. So does a light bulb. But the sun has more.

Work is easy to define. It's definition is incomprehensible at this point, and your intuition about work is most certainly wrong. The word "work" in every day speech has a meaning that is similar, but not exactly the same, to the meaning in physics. That creates real problems that needs time and effort to overcome. (I need some credit avoiding the obvious "work" pun in that sentence.)

Work in One Dimension with Constant Force

Everything is only in one dimension and any forces are constant throughout. Under these specific conditions, work can be calculated easily. Representing both force and displacement using one-dimensional vector notation, multiply them: work = (force) x (displacement). I cannot stress enough how important it is to only use this expression under the right conditions.

This is a good point to remind everyone that you use symbols to simplify writing, but you never read symbols. You only read words. Always read what the symbols represent, otherwise the symbols lose meaning. In the equation above, because it was simple, I used no symbols at all. It is unnecessary to abstract the quantities. Also, the quantities can be represented any number of ways. For example, there could be subscripts or not, force could be \(F\) or \(T\), displacement could be \(d\) or \(\Delta x\) or \(ell\). The symbols are irrelevant and infinite. The words and the physics they represent are important and unique. I may write, "\(W=Fd\)", but I would never read (nor, worse, memorize) "double-u equals eff times dee". Read "work equals force times distance" and memorize "When the system is constrained to one dimension and the force on an object is constant, the work the force does on the object is force times the distance when written in 1D vector notation."

Work in 1D with Variable Forces

If your force varies with time, \(F(t)\), you can try to break it into segments where the force is constant in time during that segment. If the force varies continuously, you can go to smaller and smaller segments, until each segment is so, we might say, infinitesimally small ... I think you see where this is going. The work for a variable force is given by: $$\int_{t_i}^{t_f }F(t)v(t)\, dt$$ where \(v(t)\) is the velocity. Remember this is still just in one dimension. Force and velocity are vectors, but in one dimension they can be represented by numbers whose value is the magnitude and sign indicates direction.

In the very common case where the force varies in space but not time, \(F(x)\) then the above can be re-written as: $$\int_{x_i}^{x_f} F(x)\, dx $$ which is convenient because we do not need to know the velocity.


Do Now!   Do these exercises immediately.

Not Now!   Do these after you start to forget the topic, say in a week.

More!   More exercises if you want. Maybe review before a test.

Required Problems



The figure below shows the "two blocks" example from the module on Newton's Third Law. The surface is frictionless. If the whole system moves 6 m, what is the work done by each force and the total work on each block?

position graph



You are pushing a 50 kg crate across the frictionless ground. You start with the crate at rest pushing with 100 N. You tire and the force decreases linearly to zero in 5 seconds. What is the work you do on the crate?

Optional Problems