Work and Power

Objective: Use W = Fd cosθ with a displacement and angle.

For a constant force acting through a straight displacement, work by that force is W = Fd cosθ, where θ is the angle between force and displacement. In the same direction, 20 N over 3 m does +60 J of work; an opposite force does negative work, and a perpendicular force does zero work in this simple model. Work is a scalar energy transfer, not a synonym for effort or tiredness. The formula applies to the component of the force along the displacement and requires consistent units. Net work is the sum of work by all forces on a chosen object. Carrying a stationary object while standing still involves no displacement of that object, even though muscles expend energy.

Objective: Calculate energy transferred per time and compare rates.

Average power is energy transferred or work done divided by elapsed time, P = ΔE/Δt = W/Δt, and is measured in watts (1 W = 1 J/s). Two devices can deliver the same energy while having different power if they take different amounts of time. A 600 J transfer in 30 s is 20 W. Power is not the amount of stored energy, and a 100 W rating alone does not determine total energy use without duration and operating conditions. For constant force and velocity along the same line, P = Fv under the stated assumptions. Specify average versus instantaneous power when rates vary.

Objective: Connect the sum of force work to ΔK.

The work–energy theorem states that net work on a particle or suitable object model equals change in translational kinetic energy: W_net = ΔK. A 3 kg object accelerating from 2 m/s to 4 m/s has ΔK = 1/2 × 3 × (4² − 2²) = 18 J, so the net work is +18 J in that model. If net work is negative, translational kinetic energy decreases, though the object may continue moving. The theorem uses work by all external forces, not just one applied force. Other internal motion and rotation may require a broader energy account. This relationship connects force over displacement with changes in speed.