Question

a 1200-kg car rolling on a horizontal surface has speed v = 65 km/h when it strikes a horizontal coiled spring and is brought to rest in a distance of 2.2 m. What is the spring stiffness constant of the spring?

Answer 1

Answer:

The stiffness constant of the spring is 80831.191 newtons per meter.

Explanation:

Given that all non-conservative forces are negligible, by the Principle of Energy Conservation we get that initial translational kinetic energy of the car is turned into final elastic potential energy of the spring. That is:

$U_{f} = K_{o}$ (1)

Where:

$U_{f}$ - Final elastic potential energy of the coiled spring, measured in joules.

$K_{o}$ - Initial translational kinetic energy, measured in joules.

By definitions of elastic potential and translational kinetic energies we expand  (1):

$\frac{1}{2}\cdot k\cdot x^{2} = \frac{1}{2}\cdot m \cdot v^{2}$ (1b)

Then, we clear the stiffness constant within the expression:

$k\cdot x^{2} = m\cdot v^{2}$

$k = m\cdot \left(\frac{v}{x} \right)^{2}$

Where:

$m$ - Mass of the car, measured in kilograms.

$v$ - Initial speed of the car, measured in meters per seconds.

$x$ - Final deformation of the coiled spring, measured in meters.

If we know that $m = 1200\,kg$, $x = 2.2\,m$ and $v = 18.056\,\frac{m}{s}$, then the stiffness constant of the spring is:

$k = (1200\,kg)\cdot \left(\frac{18.056\,\frac{m}{s} }{2.2\,m} \right)^{2}$

$k = 80831.191\,\frac{N}{m}$

The stiffness constant of the spring is 80831.191 newtons per meter.