Problem — Reading the F(x) graph of a spring

Reading a graph: $F(x)$ of a spring. The graph shows the force exerted by a spring as a function of extension: a straight line through the origin, $F = kx$, with $k = 100\;\mathrm{N/m}$. Calculate the elastic energy stored for $x = 0{,}10\;\mathrm{m}$ as the area under the curve, and check it with the formula $U = \tfrac{1}{2}kx^2$.

Solution

Data: $k = 100\;\mathrm{N/m}$, $x = 0{,}10\;\mathrm{m}$. The elastic force follows Hooke’s law $F = kx$: on the graph it is a straight line through the origin.

Method 1 — area under the curve. The work done on the spring (stored energy) is the area between the line $F(x)$ and the $x$ axis, i.e. a triangle. The maximum force reached at $x = 0{,}10\;\mathrm{m}$ is

$F_\text{max} = kx = 100\cdot 0{,}10 = 10\;\mathrm{N}$

The area of the triangle (base $= x$, height $= F_\text{max}$):

$U = \tfrac{1}{2}\cdot \text{base}\cdot \text{height} = \tfrac{1}{2}\cdot 0{,}10\cdot 10 = 0{,}5\;\mathrm{J}$

Method 2 — formula. Check with $U = \tfrac{1}{2}kx^2$:

$U = \tfrac{1}{2}(100)(0{,}10)^2 = \tfrac{1}{2}(100)(0{,}01) = 0{,}5\;\mathrm{J}$

The two methods agree:

$\ev{U = \tfrac{1}{2}kx^2 = 0{,}5\;\mathrm{J}}$

Physical reading. The formula $U = \tfrac12 kx^2$ is the area of the triangle under the line $F = kx$: geometry and algebra say the same thing. Since the force is not constant but grows linearly, we do not use the rectangle $F_\text{max}\cdot x$ (that would give double), but the triangle. The factor $\tfrac12$ in the elastic formula has precisely this geometric origin.

Collegamenti

Argomenti: Work and energy Concetti: Elastic potential energy · Elastic force and Hooke’s law Competenze: Reading graphs Oggetti: Spring