Question

A 1,508 kg car rolling on a horizontal surface has a speed of 20.8 km/hr when it strikes a horizontal coiled spring and is brought to rest in a distance of 6.87 m. What is the spring constant (in N/m) of the spring

Answer 1

Answer:

Approximately $1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}$, assuming friction between the vehicle and the ground is negligible.

Explanation:

Let $m$ denote the mass of the vehicle. Let $v$ denote the initial velocity of the vehicle. Let $k$ denote the spring constant (needs to be found.) Let $x$ denote the maximum displacement of the spring.

Convert velocity of the vehicle to standard units (meters per second):

$\begin{aligned}v &= 20.8\; {\rm km \cdot h^{-1}} \times \frac{1000\; {\rm m}}{1\; {\rm km}} \times \frac{1\; {\rm h}}{3600\; {\rm s}} \\ &\approx 1.908\; {\rm m \cdot s^{-1}}\end{aligned}$.

Initial kinetic energy (${\rm KE}$) of the vehicle:

$\begin{aligned}\frac{1}{2}\, m \, v^{2}\end{aligned}$.

When the vehicle is brought to a rest, the elastic potential energy ($\text{EPE}$) stored in the spring would be:

$\displaystyle \frac{1}{2}\, k\, x^{2}$.

By the conservation of energy, if the friction between the vehicle and the ground is negligible, the initial $\text{KE}$ of the vehicle should be equal to the ${\rm EPE}$ of the vehicle. In other words:

$\begin{aligned}\frac{1}{2}\, m \, v^{2} &= \frac{1}{2}\, k\, x^{2}\end{aligned}$.

Rearrange this equation to find an expression for $k$, the spring constant:

$\begin{aligned}k &= \frac{m\, v }{x^{2}}\end{aligned}$.

Substitute in the given values $m = 1508\; {\rm kg}$, $v \approx 1.908\; {\rm m\cdot s^{-1}}$, and $x = 6.87\; {\rm m}$:

$\begin{aligned}k &= \frac{m\, v }{x^{2}} \\ &\approx \frac{1508\; {\rm kg} \times 1.908\; {\rm m\cdot s^{-1}}}{(6.87\; {\rm m})^{2}} \\ &\approx 1.79 \times 10^{5}\; {\rm kg \cdot m \cdot s^{-2} \cdot m^{-3}}\\ &\approx 1.79 \times 10^{5}\; {\rm N \cdot m^{-1}}\end{aligned}$