Work and Power Calculator — W = F·d·cosθ and P = W/t

Real-world use cases

• Check a physics homework answer with the steps shown

  The problem says a 25 N force drags a crate 8 m at 35° above the floor — find the work. Type force, distance and angle in, read W = 25 × 8 × cos(35°) ≈ 163.8 J, and follow the four formula lines so you can write the same working on paper. When the textbook answer differs, the visible steps show exactly where a dropped cosθ or a degree-versus-radian slip crept in.

• Find the power rating a motor needs

  A conveyor must do 4500 J of lifting work every 9 seconds. Switch to power mode, enter the work and the time, and read P = 500 W. The HP line shows about 0.67 HP, so when you shop for a motor quoted in horsepower you already know the size you need rather than converting in your head at the supplier.

• Size a winch or engine from force and speed

  You know a winch must pull 1200 N of cable at a steady 0.4 m/s but you never computed the work or timed anything. Enter the force and the velocity and P = F·v gives 480 W directly. This is the fast path for steady-state jobs — towing, hauling, climbing — where speed is the quantity you actually measured.

• Show students why negative work is real

  Set the angle to 120° and the tool returns negative work with a note that the force is opposing the motion. Project it in class to make the friction-and-braking case concrete: at 180° the same force removes the full F·d of energy. Share the URL and every student reopens the exact example without retyping the numbers.

• Reverse-solve an unknown force from measured work

  A lab logged 240 J of work moving a cart 6 m, and you need the average force along the track. Pick force mode, enter the work, the distance and the 0° angle, and read F = 40 N. It is the inverse of W = F·d·cosθ solved for you, with the division step shown so the lab write-up has the full derivation.

Common pitfalls

• Forgetting the cosθ term and treating every force as if it were aligned with the motion. A 50 N pull at 60° does only half the work of the same pull at 0°. If your answer is too high, check whether you dropped the angle and computed plain F·d instead of F·d·cosθ.

• Confusing work and power, or joules and watts. Work is the total energy transferred (joules); power is how fast it is transferred (watts = joules per second). Two machines can do the same 1000 J of work, but the one that finishes in half the time has twice the power. Always note whether the question asks for energy or rate.

• Entering the angle in radians when the field expects degrees. A force at 1.05 radians is the same as 60°, but typing 1.05 into a degrees field gives cos(1.05°) ≈ 0.9998, almost no reduction, and a badly wrong answer. This tool takes degrees; convert with degrees = radians × 180 / π first.

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