Work
What physics means by work
In everyday talk, studying for an exam is hard work. In physics, the word has a much narrower meaning.
Watch a worker push a heavy crate across a warehouse floor. A force acts, and the crate moves. That is work.
Now the worker pushes on a wall with all his strength. He gets tired, but the wall does not move. In physics, he does no work on the wall.
So work needs two things: a force, and a displacement of the point where the force acts.
When the force and the displacement point the same way, work is simply force times distance. W equals F times d.
Push with 50 newtons through 4 metres and you do 200 newton metres of work. One newton metre is called one joule.
A joule is small. Lifting an apple about one metre off the table takes roughly one joule.
But what if the force is at an angle to the motion? That is the next question.
Work at an angle: the scalar product
A child pulls a toy cart with a string held at an angle θ above the ground. The cart rolls along the floor.
Split the pull into two arrows. F cos θ points along the floor. F sin θ points straight up.
Only the part along the motion does work. So W equals F d cos θ.
This combination has a name: the scalar product, or dot product, of F and d. W equals F dot d. The answer is a number with no direction.
If θ is less than 90 degrees, cos θ is positive and the work is positive. The force helps the motion.
At exactly 90 degrees, cos θ is zero, so the work is zero. Gravity does no work on a ball rolling along a level floor.
Beyond 90 degrees, cos θ is negative. Friction on a sliding box points backwards, at 180 degrees, so friction does negative work.
With components it is even quicker: F dot d equals Fx dx plus Fy dy plus Fz dz.
Try it: pull a block
- Change the angle θ and watch the dashed blue arrow, F cos θ. Only that part does work.
- Set θ to 90° and see the work drop to zero.
- Turn on the rough floor: friction does negative work, gravity and the normal force do none.
Quick check
A coolie at a railway station carries a suitcase on his head and walks along the level platform at a steady speed. How much work does his upward force on the suitcase do?
- Positive work, because he is tired
- Negative work, because the force is upward
- Zero work
- It depends on how heavy the suitcase is
Answer
C. His force on the suitcase is straight up and the suitcase moves sideways. The angle is 90°, and cos 90° = 0, so W = F d cos 90° = 0. Feeling tired is about the work done inside his muscles, not on the suitcase.
Using components
A force F = (3î + 4ĵ) N acts on a body while it moves through d = (2î − ĵ) m. How much work does the force do?
- 10 J
- 2 J
- −2 J
- 5 J
Answer
B. Use the scalar product: W = F·d = Fₓdₓ + F_y d_y = (3)(2) + (4)(−1) = 6 − 4 = 2 J. The y-part of the force acts against the y-part of the displacement, so it does negative work and cancels some of the x-part.
Summary
- Work done by a constant force:
W = F·d = F d cos θ - Scalar product in components:
A·B = AₓBₓ + A_yB_y + A_zB_z - Unit of work:
1 J = 1 N m = 1 kg m² s⁻²
- Work needs a force and a displacement. No displacement, no work.
- Only the part of the force along the displacement, F cos θ, does work. Work is a scalar.
- θ < 90°: positive work. θ = 90°: zero work. θ > 90°: negative work, like friction on a sliding box.