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

`(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{(2mgh)/(k)}`

`x = \sqrt{(2(13000)(10))/(1*10^6)}`

`x = 0.5099m`

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