Hooke’s Law: How Stiff Are Car Springs?

Problema e solução em inglês, como no livro original.

What is the force constant for the suspension system of a car, like that shown in Figure 5.41, that settles 1.20 cm when an 80.0-kg person gets in?

Figure 5.41 A car in a parking lot. (exfordy, Flickr)

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Estratégia
Consider the car to be in its equilibrium position x = 0 before the person gets in. The car then settles down 1.20 cm, which means it is displaced to a position x = −1.20×10⁻² m.

At that point, the springs supply a restoring force F equal to the person’s weight

w = mg = (80.0 kg)(9.80 m/s²) = 784 N. We take this force to be F in Hooke’s law.
Knowing F and x, we can then solve for the force constant k.
Solução
Solve Hooke’s law, F = −kx, for k.
$k = \frac{F}{x}$
Substitute known values and solve for k.
$\begin{matrix} k & = & \frac{- 784 N}{- 1.20 \times 10^{- 2} m} \\ = & 6.53 \times 10^{4} N/m \end{matrix}$

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