Q73P - University Physics with Modern Physics

Question

You are asked to design spring bumpers for the walls of a parking garage. A freely rolling 1200 kg car moving at 0.65 m/s is to compress the spring no more than 0.090 m before stopping. What should be the force constant of the spring? Assume that the spring has negligible mass. 

Step-by-Step Solution

Answer

The spring has the force constant equal to $6.3\times {10}^4 \mathrm{N}/\mathrm{m}$.

Step 1: Identification of given data

The given data can be listed below,

• The mass of the car is, $m=1200 \mathrm{kg}$
• The velocity of the car is, $v=0.65 \mathrm{m}/\mathrm{s}$
• The compression of the spring is, $d=0.90 \mathrm{m}$ 

Step 2: Concept/Significance of spring constant.

The ratio of the force applied to the spring to the compression it causes gives us the force constant of the spring. It is also known as the spring constant.

Step 3: Determination of the force constant of the spring

From the work-energy theorem, the force constant of the spring $\left(k\right)$ is obtained from the relation

$$\begin{gathered} -\frac{1}{2}k{x}^2=\frac{1}{2}m\left(v_f^2-v_0^2\right) \\ k=-\frac{m\left(v_f^2-v_0^2\right)}{x^2} \end{gathered}$$ 

Here, m is the mass of the car, x is the displacement of car, $v_f$ is the final velocity of the car and $v_o$ is the initial velocity of the car.

Substitute the given values in the above equation,

$$\begin{gathered} k=-\frac{\left(1200 \mathrm{kg}\right)\left({\left(0\right)}^2-{\left(0.65 \mathrm{m}/\mathrm{s}\right)}^2\right)}{{\left(0.090 \mathrm{m}\right)}^2} \\ =6.3\times {10}^4 \mathrm{N}/\mathrm{m} \end{gathered}$$ 

Thus, the force constant of spring is $6.3\times {10}^4 \mathrm{N}/\mathrm{m}$ .