Work
1. Work is a transfer, and the definition is an integral
Work is the energy moved into or out of a system by a force exerted over a distance. In this course the definition is written in the form that is always true, rather than the form that is true when nothing is changing:
W = ∫ F · dralong the path the point of application takes
Read it as an instruction. Cut the path into pieces short enough that the force does not change across one. On each piece take the part of the force lying along it, multiply by the length of the piece, and add up every one. That is all an integral is here.
The same curved force profile, cut into a number of slices you control. Watch the sum of the rectangles approach the area as the slices get thinner. Nothing about the physics changes; only the arithmetic gets honest.
2. The dot product does the projecting
The dot product of two vectors is the product of their magnitudes times the cosine of the angle between them, and it returns a scalar.
A · B = A B cos θ = A_x B_x + A_y B_ya scalar, from two vectors
It is exactly the tool work needs, because work asks for the part of the force that lies along the motion and nothing else. Two perpendicular vectors dot to zero, which is why a perpendicular force does no work at any point of any path.
Sweep the angle between the force and the displacement. The projection is drawn, the sign follows the cosine, and the work bar flips when the angle passes a right angle.
3. The constant-force case, which is a result and not a definition
If the force is constant in size and direction, and the path is straight, it comes outside the integral and what is left integrates to the displacement:
W = F · d = F d cos θtwo conditions, both required
That is the AP Physics 1 formula. Here it is the special case, and checking both conditions before using it is the difference between a method and a habit.
4. The work-energy theorem falls out
Start from the definition, put Newton’s second law in for the force, change the variable from position to speed, and the theorem appears in three lines:
∫ F · dr = ∫ m (dv/dt) v dt = ∫ m v dv = ½ m v_f² − ½ m v_i²
Do it once with your own pen. A theorem you have derived is one you can rebuild under pressure; a theorem you have memorized is one you can misremember.
5. Friction charges by the path
Energy dissipated by friction is the friction force times the length of the path, not the net displacement. A block that slides out and back has an integral of the displacement equal to zero and has warmed the floor the whole way.
Two routes between the same two points. Gravity is conservative and does not care which one you took. Friction is not and does.
That is also the cleanest argument for why friction has no potential energy function. A potential energy is a function of position, so its change around any closed path has to be zero. Friction takes energy on every closed path, so no such function can exist.
Check yourself
1. A force F = 3x newtons acts along the x axis from x = 0 to x = 2.0 m. Find the work it does.
W = ∫₀² 3x dx = (3/2)x² |₀² = 6.0 J. The formula with the cosine would have given the wrong answer here, because the force is not constant.
2. Find A · B for A = 3i + 4j and B = 2i − j, then find the work done by F = 5i + 2j N through a displacement d = 4i m.
A · B = (3)(2) + (4)(−1) = 2. The work: (5)(4) + (2)(0) = 20 J. The j component of the force contributed nothing, because the displacement had no j component.
3. Derive W = Fd cos θ from the integral definition, and state both conditions you used.
With F constant it comes outside the integral, leaving F · ∫ dr. With the path straight, ∫ dr is the displacement d, so W = F · d = Fd cos θ. Constant force, straight path. Break either and the result goes with it.
4. State the work done by the magnetic force on a moving charged particle, and say why.
Zero, always. The magnetic force is perpendicular to the velocity by construction, so F · dr vanishes at every point of the path and the integral of zero is zero. A magnetic field can bend a particle path and can never change its speed.
Topic 3.3, Potential Energy. Integrate a conservative force and you get a potential energy. Differentiate the potential energy and you get the force back. This course asks for both directions.