Question

A 13,000 N car starts at rest and rolls down a hill from a height of 10 m. It then moves across a level surface and collides with a light spring-loaded guardrail. Neglecting any losses due to friction, find the maximum distance the spring is compressed. Assume a spring constant k of 1.0x106 N/m (0.50 m)

Answer 1

To solve this problem it is necessary to apply the concepts related to the conservation of kinetic and potential energy, that is to say

$KE = PE$

Where,

KE = Kinetic Energy

PE = Potential Energy (in a Spring)

Using the expression we have that

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

Here,

k = Spring constant

x = Displacement

m = mass

g = Gravitational acceleration

h = Height

Re-arrange to find the displacement,

$x = \sqrt{\frac{2mgh}{k}}$

$x = \sqrt{\frac{2(13000)(10)}{1*10^6}}$

$x = 0.5099m$

Therefore the maximum distance the spring is compressed around to 0.5m