Question

What should the spring constant k of a spring be in order for it to bring a 1480 kg car to rest from a speed of 91.0 km/h such that the passengers undergo a maximum acceleration of 5g?

Answer 1

Answer:

k = 5572 N/m

Explanation:

Energy stored in the spring:

1) $U=\frac{1}{2}kx^{2}$

Kinetic energy of the car:

2) $K = \frac{1}{2}mv^{2}$

Spring force:

3) $F_S=-kx$

Combining equation 1 and 2 and solving for k:

$\frac{1}{2}kx^{2}=\frac{1}{2}mv^{2}$

4) $k = m(\frac{v}{x})^{2}$

Using equation 3 and solve for x:

5) $x=-\frac{m}{k}a$

Combining equation 4 and 5:

6) $k=m\frac{a^2}{v^2}=25m\frac{g^2}{v^2}$