Work and the Work-Energy Theorem
Key Takeaways
- Work done by a constant force is W = Fd cos θ, where θ is the angle between the force vector and the displacement vector — only the component of force parallel to displacement does work
- The Work-Energy Theorem states that the net work done on an object equals its change in kinetic energy: W_net = ΔKE = ½mv_f² − ½mv_i²
- Mechanical advantage lets a machine trade force for distance, or distance for force, but an ideal frictionless machine never reduces the total work required to move a load
- Conservative forces (gravity, the spring/elastic force) do path-independent work and store energy as recoverable potential energy; non-conservative forces (friction, air resistance) dissipate energy as heat and are path-dependent
- A force applied perpendicular to an object's displacement (θ = 90°) does zero work on that object, even if the force is large — a frequently tested MCAT trap
Work Done by a Constant Force
In physics, work (W) is done on an object when a force causes a displacement of that object. For a constant force, work is calculated as:
W = Fd cos θ
where F is the magnitude of the force (in newtons, N), d is the magnitude of the displacement (in meters, m), and θ is the angle between the force vector and the displacement vector. Work is a scalar quantity — it has magnitude but no direction — and is measured in joules (J), where 1 J = 1 N·m = 1 kg·m²/s².
The cosine term is the single most tested feature of this equation on the MCAT. It isolates the component of force that actually acts along the direction of motion.
| Angle θ between F and d | cos θ | Result |
|---|---|---|
| 0° (force parallel to motion) | 1 | Maximum positive work; the force fully contributes |
| 37° (common MCAT angle) | ≈ 0.8 | About 80% of Fd contributes as work |
| 90° (force perpendicular to motion) | 0 | Zero work, regardless of the force's magnitude |
| 180° (force opposite to motion) | −1 | Maximum negative work; the force removes kinetic energy |
The 90° row is a favorite trap. A force can be large and clearly "doing something" — the normal force supporting your body weight as you walk, or the centripetal force holding a satellite in circular orbit — yet do zero work, because it acts perpendicular to displacement at every instant. Likewise, when you carry a bag of groceries at constant height across a room, the upward force you apply does zero work on the bag because the displacement is horizontal, even though your arm gets tired — a physiological (metabolic) cost that is entirely separate from mechanical work done on the object.
Worked example: A physical therapist applies a 50 N force at 37° above the horizontal to pull a patient on a sliding board 4 m along a smooth track (use cos 37° ≈ 0.8):
W = (50 N)(4 m)(0.8) = 160 J
Only the 40 N horizontal component (50 × 0.8) does work along the displacement; the 30 N vertical component (50 × 0.6) does zero work on the horizontal motion.
Mechanical Advantage and Ideal Machines
Simple machines — levers, pulleys, inclined planes, and wheel-and-axle systems — do not create energy. Instead, they redistribute the same total work between force and distance. Mechanical advantage (MA) is the ratio of the output (load) force to the input (effort) force:
MA = F_load / F_effort
In an idealized, frictionless machine, the work put in equals the work delivered out: F_effort × d_effort = F_load × d_load. A ramp that is 5 m long and rises 1 m, for example, lets you push a load up using only about one-fifth of the force required to lift it straight up — but you must push it five times farther, so the total work is unchanged. Real machines have efficiency below 100% because friction, a non-conservative force, converts some of the input work into heat, so the actual output work is always somewhat less than the input work.
In rehabilitation and biomechanics, ramps and lever-based assistive devices reduce required force for patients with limited strength, but they still require roughly the same total work (plus friction losses). Passages may ask whether a longer ramp "saves energy" — for an ideal frictionless case the answer is no; it only reduces force at the expense of distance.
The Work-Energy Theorem
The Work-Energy Theorem connects the work done on an object to its motion:
W_net = ΔKE = ½mv_f² − ½mv_i²
The net work — the sum of the work done by every force acting on the object, equivalently the work done by the net force — equals the object's change in kinetic energy. This is one of the most useful shortcuts on the MCAT because it allows you to find a final speed without solving for acceleration or applying kinematics equations directly. If net work on an object is positive, its kinetic energy (and speed) increases; if negative, kinetic energy decreases; if zero, speed is unchanged, even if individual forces are separately doing positive and negative work that happens to cancel.
A key distinction: the work done by one force (say, tension in a rope) is not the same as net work unless that is the only force acting on the object. When multiple forces act together — tension, gravity, friction, a normal force — you must either sum the work done by each force individually or first find the net force before applying W = Fd cos θ.
Worked example: A 2 kg object initially moving at 3 m/s is acted on by a net force that does +32 J of work on a frictionless surface.
Initial KE = ½(2)(3)² = 9 J Final KE = 9 J + 32 J = 41 J ½(2)v_f² = 41 → v_f² = 41 → v_f = √41 ≈ 6.4 m/s
The sign of work also has a direct physiological parallel worth remembering for the MCAT: a muscle shortening while producing tension (a concentric contraction, such as the biceps lifting a weight) does positive work on the load, while a muscle lengthening under tension (an eccentric contraction, such as the same biceps slowly lowering that weight) does negative work — the load's gravitational force is doing positive work on the muscle-tendon unit instead. Both contraction types cost metabolic energy, which is why lowering a heavy object under control is still tiring even though gravity, not muscle, is doing the positive mechanical work on the load.
Conservative vs. Non-Conservative Forces
Forces fall into two categories based on how they handle energy.
Conservative forces — gravity and the spring (elastic) force are the two conservative forces tested on the MCAT — do work that depends only on an object's initial and final position, not on the path taken between them. Move an object around a closed loop under a conservative force alone, and the net work done is exactly zero; all the energy "borrowed" while moving away from the starting point is fully returned. Because of this path-independence, conservative forces can be associated with a potential energy function, the subject of the next section.
Non-conservative forces — friction, air resistance, and applied forces such as muscle contraction — are path-dependent: a longer or rougher path costs more energy. These forces convert usable mechanical energy into heat (and sometimes sound), which cannot be recovered as mechanical work. This is why no real machine is 100% efficient, and it is also why muscle contraction is metabolically expensive: only a fraction of the chemical energy released by adenosine triphosphate (ATP) hydrolysis becomes useful mechanical work at a joint, with the remainder lost as heat — a theme this guide returns to when covering thermodynamics and bioenergetics.
MCAT traps for work and energy:
- Computing Fd without cos θ when force is at an angle
- Claiming the normal force does work during horizontal walking (θ = 90° → W = 0)
- Equating work by one force with net work when friction or gravity also acts
- Assuming machines reduce total work rather than trading force for distance
- Treating friction work as recoverable potential energy
A worker pushes a crate across a horizontal floor by applying a force of 50 N directed 37° above the horizontal (use cos 37° ≈ 0.8). If the crate moves 4 m, how much work does the applied force do on the crate?
A 2 kg object initially moving at 3 m/s is acted on by a net force that does +32 J of work as the object travels across a frictionless surface. What is the object's final speed?
While walking at constant height, a person carries a 10 N medical bag horizontally for 20 m. How much mechanical work does the upward force supporting the bag do on the bag during this walk?