Work: Study Guide (Physics C: Mechanics 3.2)

How this differs from the Physics 1 version. Physics C: Mechanics and Physics 1 are separate courses that both have a Topic 3.2 called Work. This guide is the calculus-based one. It defines work as an integral of a dot product along a path, integrates forces that change with position, and derives the work–energy theorem from Newton’s second law. The algebra-based treatment is in the Physics 1 study guide; do not mix the two when you revise. This topic follows Topic 3.1, Translational Kinetic Energy.

Work is an energy transfer

Choose a system first. A force exerted on that system by something outside it can transfer energy into or out of the system as the point where the force acts moves. That transfer is work.

• W > 0: energy goes into the system.
• W < 0: energy leaves the system.
• W = 0: no energy is transferred by that force.

Work is a scalar, measured in joules: 1 J = 1 N·m. A force does no work if its point of application does not move, however large the force is.

The dot product and constant forces

Work uses the dot product of two vectors. For vectors A and B at angle θ:

A · B = AB cos θ = A_xB_x + A_yB_y + A_zB_z

The result is a scalar. For a constant force F whose point of application undergoes displacement d:

W = F · d = Fd cos θ

• θ < 90°: W > 0. θ = 90°: W = 0. 90° < θ ≤ 180°: W < 0.
• F cos θ is the component of the force parallel to the displacement, F‖. So W = F‖d. Only that component changes the system’s energy.
• The perpendicular component does no work. It can change the direction of motion without changing K. The tension on a ball in uniform circular motion (Topic 2.10) is always perpendicular to the velocity, so it does zero work and the speed stays constant.

Component example. F = (3.0, 4.0) N acts while its point of application moves d = (2.0, −1.0) m. W = (3.0)(2.0) + (4.0)(−1.0) = 2.0 J. You never needed the angle; with magnitudes 5.0 N and 2.24 m, it is about 80°.

Variable forces: integrate along the path

If the force changes along the path, split the path into tiny displacements dr. Over each, the force is nearly constant, so the work is F · dr. Add them all:

W = ∫ₐᵇ F · dr

The integral runs along the actual path from a to b. In one dimension, with the force and motion along x, this becomes W = ∫ F_x dx.

Graph meaning. Work is the signed area under a graph of F‖ against position. Area above the axis is positive work; area below is negative.

Spring force. An ideal spring (Topic 2.8) exerts F_s,x = −kx, with x measured from the relaxed length. The work it does as the end moves from x_i to x_f is

W_s = ∫ (−kx) dx = −½k(x_f² − x_i²)

Stretching a 150 N/m spring from 0 to 0.10 m, the spring does −0.75 J of work on whatever pulls it. Letting it return to 0, the spring does +0.75 J. Round the full trip, the total is zero.

Net work and the work–energy theorem

The net work is the sum of the work done by every force on the object, or the work done by the net force. Newton’s second law turns it into a change in kinetic energy. In one dimension, use the chain rule as in Topic 1.2:

F_net,x = m dv_x/dt = m (dv_x/dx)(dx/dt) = m v_x dv_x/dx

So F_net,x dx = m v_x dv_x. Integrate from the initial to the final state:

W_net = ∫ F_net,x dx = ½mv_f² − ½mv_i² = ΔK

This is the work–energy theorem. In three dimensions the same steps use F_net · dr = m (dv/dt) · v dt = d(½mv²). It links straight back to Topic 3.1: dK/dt = F_net · v.

Worked example 1: a variable push

Question. Take +x along a level, frictionless track. A 2.0 kg cart starts from rest at x = 0. A launcher pushes it with F_x = (12 N) − (3.0 N/m²)x² from x = 0 to x = 2.0 m, where the push falls to zero. Find the work done and the cart’s speed at x = 1.0 m and at x = 2.0 m.

1. Gravity and the normal force are perpendicular to the displacement, so they do no work. The push is the only force doing work.
2. Work from 0 to 2.0 m: W = ∫₀² (12 − 3.0x²) dx = [12x − x³]₀² = 24 − 8.0 = 16 J.
3. Work–energy theorem: ½(2.0)v² = 16, so v = 4.0 m/s at x = 2.0 m.
4. At x = 1.0 m: W = 12 − 1.0 = 11 J, so v = √(2 × 11 ÷ 2.0) = 3.3 m/s.

Check. The areas in Figure 1 are 11 J and 5 J; they add to 16 J. Using the starting force for the whole distance, 12 N × 2.0 m = 24 J, overestimates the work because the force falls. The speed keeps rising all the way to x = 2.0 m, because F_x stays positive.

Worked example 2: several constant forces on a ramp

Question. Take +x down the slope. A 4.0 kg box slides 3.0 m down a ramp inclined at 30°. The coefficient of kinetic friction is 0.20. The box starts at 1.0 m/s. Find the work done by each force and the final speed.

1. Gravity. The parallel component is mg sin 30° = 19.6 N, down the slope. W_g = (19.6)(3.0) = +59 J (58.8 J). Check with components, +y up: F_g = (0, −39.2) N and d = (2.60, −1.50) m, so W_g = (−39.2)(−1.50) = 58.8 J. Only the vertical drop of 1.5 m matters.
2. Normal force. Perpendicular to the ramp, so W_N = 0. Its size is mg cos 30° = 33.9 N.
3. Friction. f = 0.20 × 33.9 = 6.79 N, up the slope, opposite to d. W_f = −(6.79)(3.0) = −20 J (−20.4 J).
4. Net work: 58.8 + 0 − 20.4 = +38 J (38.4 J).
5. Final speed: K_i = ½(4.0)(1.0)² = 2.0 J. K_f = 2.0 + 38.4 = 40.4 J. v = √(2 × 40.4 ÷ 4.0) = 4.5 m/s.

Interpretation. Gravity transfers energy into the box; friction takes 20 J out. That 20 J does not vanish: it becomes thermal energy and some sound in the box and ramp. This course tracks the mechanical energy only.

Conservative and nonconservative forces

Some forces do the same work whatever path the object takes between two points. These are conservative forces.

• The work depends only on the start and end configurations of the system.
• Round any closed path, the total work is zero.
• Only conservative forces have a potential energy linked to them. Gravity and ideal springs are the examples you know; Topic 3.3 builds potential energy from them.

Other forces do work that depends on the path. These are nonconservative. The common ones are kinetic friction and air resistance. A longer path means more work done against them. The mechanical energy removed by friction is usually taken as friction force × path length.

Worked example 3: path dependence

Question. A 2.0 kg block is pushed at constant speed across a level floor from A to B (Figure 2), with μ_k = 0.25. Find the work done by friction along each path, and round the closed trip A → B → A along path 1. Then compare with the work done by gravity when the same block is lifted 1.5 m onto a shelf by any route.

1. Friction force: f = μ_k mg = 0.25 × 2.0 × 9.8 = 4.9 N, always opposite to the motion.
2. Path 1 (5.0 m): W_f = −4.9 × 5.0 = −24.5 J. Path 2 (7.0 m): W_f = −34.3 J. Different paths, different work: friction is nonconservative.
3. Closed trip along path 1 and back: friction still opposes the motion on the way back. W_f = −4.9 × 10 = −49 J, not zero.
4. Lifting by 1.5 m: gravity’s work is W_g = −mgΔy = −2.0 × 9.8 × 1.5 = −29 J (−29.4 J), whether the block goes straight up, up a ramp, or along a zigzag. Bring it back down and gravity does +29.4 J: zero round the loop. Gravity is conservative.

Object or system: where does the force act?

The work done by a force uses the displacement of the point where the force acts, not automatically the displacement of the centre of mass.

• If the point of application and the centre of mass move the same distance (a crate pushed by a rigid rod), the system can be modelled as an object. The work done by the external force then shows up only as a change in K.
• If they move differently, the system’s internal configuration can change. When you push off a wall, the wall’s force on your hands does no work, because your hands do not move while the force acts. Your centre of mass still speeds up. The energy came from inside you (chemical energy in your muscles), not from the wall. You cannot treat yourself as a single object here.

Common misconceptions

• “A large force always does a lot of work.” No displacement of the point of application, no work.
• “Work has a direction.” It is a scalar. Its sign shows whether energy goes in or out.
• Using F × d for a varying force. Integrate, or find the area under the F‖–x graph (Worked example 1).
• “The normal force always does no work.” It does none when it is perpendicular to the displacement, as on a fixed ramp. The floor of a lift that is moving upward pushes up on you through a displacement, so it does positive work on you.
• “Centripetal force does work because it acts all the way round.” It is perpendicular to the velocity at every instant, so W = 0.
• “Friction’s work only depends on start and end points.” It depends on the path length (Worked example 3).
• Applying W_net = ΔK to only one force. The theorem uses the work done by all forces.

Where this leads

Topic 3.3, Potential Energy turns the work done by conservative forces into stored energy, using ΔU = −W. Topic 3.4 combines work, kinetic energy and potential energy into energy conservation, and Topic 3.5 treats the rate of doing work as power. Try the practice questions now, then use the revision notes and the checklist. You can also return to the course roadmap.