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Wize AP Physics C: Mechanics Textbook > Unit 3: Work, Energy, and Power (14-17%)

Mechanical Work (Calculus-Based / Using Integrals)

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Mechanical Work


Mechanical work is defined as the dot product of displacement vector and force vector. For a constant force applied over the object in Δr⃗\Delta\vec{r} distance, the work done by the force is:

W=F⃗. Δr⃗ =FΔrcos⁡(θ)\boxed{W=\vec{F}.\ \Delta\vec{r}\ =F\Delta r \cos(\theta)}




where θ\thetais the angle between force direction and direction of motion of the object.



For a varying force, work is defined as:

W=∫r1r2F⃗ . dr⃗\boxed{W=\int_{r_1}^{r_2}\vec{F}\ .\ d\vec{r}}


Wize Tip
If the curve of force as a function of x is given for a force along the motion, the area under the curve is the work done by the force.


The SI unit of work is the Joule.1 J=1N⋅m=1 kg⋅m2s21\ J=1N\cdot m=1\ \frac{kg\cdot m^2}{s^2}

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Mechanical Work Sign

Work is a scalar quantity and can be either positive, negative or zero:
  1. If the applied force F (and the displacement d are in the same direction, work done is positive – e.g: pushing block forward
  2. If F and d are in opposite directions (cos 180° = −1), work done is negative – e.g. applying brakes in a moving car
  3. Work is zero if either
  • displacement is zero, so the work done is zero – e.g. pushing against a wall OR
  • θ = 90°, so cos θ = 0 (force is perpendicular to displacement) – e.g. you are carrying a box and moving horizontally (no work done against the force of gravity)

Three cases:





Determine the work done by the variable force F(x) = 3x2 from x=0 to 2m.



W = ∫ dW = ∫ F dx = ∫ 3x2dx = x3 (evaluate from 0 to 2) = 8 - 0 = 8 Nm
Plotting the function from 0 to 2m, the Work Done is the area under the curve.


An object moving along x-axis is acted upon by a variable force F−xF-x . How much work does the force do on the object as the object moves from x=0 to x=16 m?



Solution: