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

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

Step-by-step 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)} * \frac{1000\; {\rm m}}{1\; {\rm km}} * \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}(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 (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}(1)/(2)\, m \, v^(2) &= (1)/(2)\, k\, x^(2)\end{aligned}`.

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

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