Calculate work done by a force at an angle to the displacement and the power at which energy is transferred

Learning goal: Calculate work with W = FΔx cos θ when the force and the displacement are at an angle, recognize when a force does no work, use free body diagrams to find the forces involved, and calculate average power and the instantaneous power delivered to an object moving at constant velocity.

Work when force and displacement are not aligned. Work is the transfer of energy to or from an object by a force acting through a displacement. In Physical Science you calculated it only when the force pointed along the motion. The general expression for a constant force is W = FΔx cos θ, where F is the magnitude of the force, Δx is the magnitude of the displacement and θ is the angle between the direction of the force and the direction of the displacement. Only the component of the force along the displacement, F cos θ, transfers energy. The unit is the joule (1 J = 1 N·m). Three cases follow directly from the cosine. When θ = 0° the force helps the motion and the work is positive: energy is transferred to the object. When θ = 180° the force opposes the motion, cos θ = −1, and the work is negative: energy is transferred out of the object, as friction does. When θ = 90° the cosine is zero, so the force does no work and transfers no energy, however large it is. This is why the normal force on a box sliding across a level floor does no work, and why the tension in a string keeping a ball in uniform circular motion does no work: the force is always perpendicular to the motion, so the ball's speed never changes.

Worked model: a rope at an angle. A child pulls a sled 12 m across level snow with a rope that makes a 30° angle with the ground, keeping a steady 40 N tension in the rope. Work done by the rope on the sled: W = (40 N)(12 m)(cos 30°) = (40)(12)(0.866) = 416 J. The rope's vertical component, 40 sin 30° = 20 N, lifts part of the sled's weight off the snow but does no work, because the sled does not move vertically. If the child instead pulled horizontally with the same 40 N, the work would be the full 480 J, so pulling at an angle transfers less energy per meter unless it also reduces friction, which is the next question.

Above, below or horizontal? Use a free body diagram. The angle of a pull changes the normal force, and through it the friction, so the best angle depends on whether friction is present. Take a 10 kg box on a level floor with a coefficient of kinetic friction of 0.30, pulled 5.0 m by a 50 N force. Draw the free body diagram: weight mg = 98 N down, the normal force N up, the applied force at its angle, and kinetic friction f = 0.30 × N opposing the motion. Vertical forces balance because the box does not leave the floor. Horizontal pull: N = 98 N, f = 29.4 N; work by the pull = 250 J, work by friction = −(29.4)(5.0) = −147 J, net work = 103 J. Pull 25° above horizontal: the upward component 50 sin 25° = 21.1 N reduces the normal force to 98 − 21.1 = 76.9 N, so f = 23.1 N; work by the pull = (50)(5.0)(cos 25°) = 227 J, work by friction = −115 J, net = 111 J. Pull 25° below horizontal: the downward component raises the normal force to 119.1 N, f = 35.7 N; work by the pull is again 227 J, friction −179 J, net = 48 J. Conclusion, stated with its conditions: without friction, a horizontal pull transfers the most energy (250 J versus 227 J), because all of the force acts along the motion; with friction present, a pull angled above the horizontal can transfer more net energy (111 J versus 103 J) because it lightens the normal force and so reduces the friction, while a pull below the horizontal is worst on both counts. Whether the upward angle wins depends on the coefficient of friction and the angle, so the calculation must be done, not assumed.

Power: the rate of energy transfer. Power is how fast energy is transferred or transformed: P = ΔE/Δt, or for work, P = W/Δt. Power is a scalar; it has a size but no direction. Its unit is the watt (W), equal to one joule per second, and 1 kW = 1000 W. Average power is the total energy transferred divided by the total time. Instantaneous power is the rate at one moment. A force F acting along an object's velocity v delivers power P = Fv at that instant, whether or not the object is speeding up, because in a very short time Δt the force acts through a displacement vΔt; if the force is at an angle θ to the velocity, only its component F cos θ counts. In this course, calculations of instantaneous power are limited to objects moving at constant velocity, where the force and the speed are steady and the instantaneous power equals the average power. Worked model: a 55 kg student climbs a flight of stairs 4.0 m high in 6.0 s. Work done against gravity: W = mgΔh = (55)(9.8)(4.0) = 2156 J ≈ 2.2 kJ. Average power: P = 2156 J / 6.0 s = 359 W ≈ 360 W. A heavier student who takes the same time develops more power; the same student climbing in 12 s develops half the power but does the same work.

Misconception: power measures strength or effort. Standing still holding a heavy box is tiring, but the box's displacement is zero, so the work done on it and the power delivered to it are both zero; the effort is internal to your muscles. Similarly, carrying the box across a level floor at constant velocity does no work on the box, because the supporting force is vertical and the displacement is horizontal (θ = 90°). Power in physics is always energy transferred per unit time, and it can be high with little effort (a falling rock delivers energy quickly) or zero with great effort.

Investigation: the rate at which a student can do work. Materials: a flight of stairs with a handrail, a meter stick or tape, a stopwatch, a bathroom scale. Steps: measure the vertical height of the flight by measuring one step and counting steps; record each participant's mass; each participant climbs briskly, one at a time, holding the rail, while a partner times the climb from the moment the first foot leaves the floor to the moment both feet are on the top landing. Record mass, height, time; calculate W = mgΔh and P = W/Δt. Repeat twice and average. A result that supports the model: heavier or faster climbers show larger power; the same person's work is the same on every trial while power changes with time. Discuss reasons for differences: mass, time, technique and the precision of the stopwatch (a 0.3 s reaction-time error on a 6 s climb is a 5 % error in power). What the calculation cannot show: the energy the body spent on internal processes and on heating; the work against gravity is only part of the energy the climber used. Safety: walk, do not run; no one climbs while feeling unwell; keep the stairwell clear.

Worked example 1 — the sled and its forces. Rope tension 40 N at 30° above horizontal, displacement 12 m along level snow. Work by the rope: W = FΔx cos θ = (40)(12)(0.866) = 416 J. Work by the normal force and by gravity: both act vertically while the displacement is horizontal, so θ = 90° and each does zero work. If friction on the sled is 20 N, the work by friction is (20)(12)(cos 180°) = −240 J, and the net work on the sled is 416 − 240 = 176 J, which by the work-energy relation is the gain in the sled's kinetic energy.

Worked example 2 — choosing the pull angle with friction. For the 10 kg box (μ = 0.30, F = 50 N, Δx = 5.0 m): free body diagram gives N = 98 N (horizontal pull), 76.9 N (25° above) or 119.1 N (25° below). Friction f = μN = 29.4 N, 23.1 N and 35.7 N. Work by the pull: 250 J, 227 J, 227 J. Work by friction: −147 J, −115 J, −179 J. Net work transferred to the box: 103 J, 111 J, 48 J. So with friction present, the 25° upward pull transfers the most energy; without friction, the horizontal pull (250 J) would.

Worked example 3 — average power on the stairs. Mass 55 kg, height 4.0 m, time 6.0 s. W = (55)(9.8)(4.0) = 2156 J. P = 2156/6.0 = 359 W. A 70 kg student who climbs the same flight in 7.0 s: W = 2744 J, P = 392 W. The heavier student did more work and developed more power despite taking longer.

Worked example 4 — instantaneous power at constant velocity. A truck tows a trailer along a level road at a constant 20 m/s, with a steady 600 N tension in the tow bar. Because the velocity is constant, the tension is balanced by the trailer's resistive forces, and the power delivered by the tow bar is P = Fv = (600 N)(20 m/s) = 12 000 W = 12 kW. In 60 s the tow bar transfers (12 000)(60) = 7.2 × 10⁵ J, all of which ends up as thermal energy in the tires, bearings and air, since the trailer's kinetic energy is not changing.

Representation task: three free body diagrams, one table. For a 6.0 kg crate pulled 4.0 m across a level floor (μ = 0.25) by a 30 N force, draw free body diagrams for the force applied horizontally, 20° above horizontal and 20° below horizontal. For each, calculate the normal force, the friction, the work by the pull, the work by friction and the net work, and enter them in a table with units. Then write one sentence stating which angle transfers the most energy with friction and which would without friction, and why they differ.

Independent performance: a power investigation report. Carry out the stair-climb investigation with at least three participants and write it up. Success criteria: height measured to the nearest centimeter with the method stated; masses and times in a table with SI units; work and average power calculated for each trial with one calculation shown in full; an explanation of at least two reasons why similar tasks gave different power values; an estimate of the percent uncertainty in power caused by the stopwatch; a paragraph explaining why walking along the level corridor to the stairs at constant velocity did no work on your body's center of mass even though it took effort; and a calculation of the instantaneous power a 600 N pull delivers to an object moving at a constant 1.5 m/s (900 W), with the condition under which P = Fv applies.