Question

A 1300 kg car starts at rest and rolls down a hill from a height of 10.0 m. It then moves across a

level (frictionless) surface and collides with a light spring-loaded guardrail with a spring constant
k of 1.0 x 106 N/m. Find the maximum distance the spring is compressed.

Answer 1

0.51 m

Step-by-step explanation:

Using the principle of conservation of energy, change in potential energy equals to the change in kinetic energy of the spring.

Kinetic energy, KE=½kx²

Where k is spring constant and x is the compression of spring

Potential energy, PE=mgh

Where g is acceleration due to gravity, h is height and m is mass

Equating KE=PE

mgh=½kx²

Making x the subject of formula

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

Substituting 9.81 m/s² for g, 1300 kg for m, 10m for h and 1000000 for k then

`x=\sqrt \frac {2*1300*9.81*10}{1000000}=0.50503465227646m\\x\approx 0.51 m`