Section 3.2: Work by Variable Forces
When the force acting on an object is not constant, work must be computed by dividing motion into small intervals or using calculus. The general definition of work for a variable force is given by an integral.
\( W = \int_{x_i}^{x_f} F(x) \, dx \)
- \( F(x) \): Force as a function of position
- \( x_i, x_f \): Initial and final positions
The integral sums the infinitesimal work \( dW = F(x)\,dx \). Graphically, the work is the area under the force vs. displacement curve.
Example 1
A force varies with position as \( F(x) = 2x \) N, where \( x \) is in meters. Find the work done in moving a particle from \( x = 0 \) m to \( x = 4 \) m.
\( W = \int_0^4 2x \, dx = [x^2]_0^4 = 16 \, \text{J} \)
Example 2
A spring with spring constant \( k = 200 \, \text{N/m} \) is stretched from equilibrium \( x=0 \) to \( x=0.1 \, \text{m} \). Find the work done.
\( F(x) = kx \)
\( W = \int_0^{0.1} kx \, dx = \tfrac{1}{2}kx^2 \big|_0^{0.1} \)
\( W = 0.5 \times 200 \times (0.1)^2 = 1 \, \text{J} \)
Practice Problems
- A force \( F(x) = 5x \) N acts from \( x=0 \) to \( x=2 \) m. Find the work done.
- A spring with \( k = 150 \, \text{N/m} \) is compressed 0.2 m. Find the work done.
- A particle experiences force \( F(x) = 10 - 2x \). Compute the work from \( x=0 \) to \( x=4 \) m.
- A variable force \( F(x) = 4x^2 \) acts from \( x=1 \) to \( x=3 \). Find the work done.
- A spring with constant \( k = 300 \, \text{N/m} \) is stretched 0.05 m. Compute the work done.