Mechanical work measures how much energy a force transfers during a displacement. Work is a scalar quantity, so positive
work adds mechanical energy, while negative work removes mechanical energy.
Work by a constant force
For a constant force acting at an angle to the displacement:
\[
W=\vec F\cdot\vec d=Fd\cos\theta
\]
Here \(F\) is the force magnitude, \(d\) is displacement, and \(\theta\) is the angle between the force and displacement.
Only the parallel component of the force does work:
\[
F_{\parallel}=F\cos\theta
\]
\[
W=F_{\parallel}d
\]
If \(0^\circ\le \theta<90^\circ\), the work is positive. If \(\theta=90^\circ\), the work is zero. If
\(90^\circ<\theta\le180^\circ\), the work is negative.
Work by several constant forces
Work contributions add algebraically:
\[
W_{\mathrm{total}}=\sum_i W_i=\sum_i F_i d\cos\theta_i
\]
This is useful when an applied force, friction, tension, gravity, and other forces act over the same displacement.
Variable force and graph area
When force changes with position, the work is the signed area under the force-displacement graph:
\[
W=\int_{x_0}^{x_1}F(x)\,dx
\]
Area above the \(x\)-axis gives positive work. Area below the \(x\)-axis gives negative work.
Graphical trapezoid method
If the graph is given by points and the force is approximated by straight line segments, the work is a sum of trapezoids:
\[
W\approx
\sum_{i=1}^{n-1}
\frac{F_i+F_{i+1}}{2}(x_{i+1}-x_i)
\]
This is the numerical area under the \(F\)-versus-\(x\) curve.
Numerical integration
If the force is entered as a function \(F(x)\), the calculator evaluates:
\[
W=\int_{x_0}^{x_1}F(x)\,dx
\]
For smooth functions, Simpson's rule gives an accurate numerical estimate:
\[
\int_a^b F(x)\,dx
\approx
\frac{h}{3}
\left[
F(x_0)+F(x_n)+4\sum_{\mathrm{odd}}F(x_i)+2\sum_{\mathrm{even}}F(x_i)
\right]
\]
Work-energy theorem
Net work is connected to kinetic energy by:
\[
W_{\mathrm{net}}=\Delta K
\]
\[
W_{\mathrm{net}}=\frac12mv_f^2-\frac12mv_i^2
\]
If the object starts from rest and all net work becomes kinetic energy:
\[
v_f=\sqrt{\frac{2W_{\mathrm{net}}}{m}}
\]
Worked example
Suppose a constant force of \(50\ \mathrm{N}\) acts at \(30^\circ\) over \(8\ \mathrm{m}\), and a variable force
\(F(x)=10x\) acts from \(x=0\) to \(x=4\ \mathrm{m}\).
Constant-force work.
\[
\begin{aligned}
W_1
&=Fd\cos\theta\\
&=(50)(8)\cos30^\circ\\
&\approx346.4\ \mathrm{J}
\end{aligned}
\]
Variable-force work.
\[
\begin{aligned}
W_2
&=\int_0^4 10x\,dx\\
&=5x^2\Big|_0^4\\
&=80\ \mathrm{J}
\end{aligned}
\]
Total work.
\[
W_{\mathrm{total}}=346.4+80=426.4\ \mathrm{J}
\]
Formula summary
| Concept |
Formula |
Meaning |
| Constant force |
\(W=Fd\cos\theta\) |
Dot-product work by one constant force. |
| Parallel component |
\(F_{\parallel}=F\cos\theta\) |
Only this component does work along the displacement. |
| Several forces |
\(W_{\mathrm{total}}=\sum_iF_id\cos\theta_i\) |
Constant-force contributions add algebraically. |
| Variable force |
\(W=\int_{x_0}^{x_1}F(x)\,dx\) |
Signed area under the force-displacement curve. |
| Trapezoid graph area |
\(W\approx\sum\frac{F_i+F_{i+1}}{2}(x_{i+1}-x_i)\) |
Area from piecewise-linear graph points. |
| Work-energy theorem |
\(W_{\mathrm{net}}=\Delta K\) |
Net work changes kinetic energy. |
| Speed from rest |
\(v=\sqrt{2W/m}\) |
Speed if positive net work becomes kinetic energy. |
Work is signed energy transfer: direction matters for constant forces, and graph area matters for variable forces.
