Question

An ore car of mass 43000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 14 m lower vertically, is a horizontally situated spring with constant 5.9 × 105 N/m. The acceleration of gravity is 9.8 m/s 2 . Ignore friction. How much is the spring compressed in stopping the ore car? Answer in units of m

Answer 1

Answer:

x = 4.4719 m

Explanation:

For answer this we will use the law of the conservation of energy, where:

$E_i = E_f$

First, we will call:

$E_i$: the car in rest

$E_f$: when the spring is compressed

so:

$E_i = E_f$

$Mgh= \frac{1}{2}KX^2$

where M is the mass of the car, g the gravity, h the altitude, K is the constant of the spring and X is the spring compressed in stopping the ore car. So, replacing values, we get:

$(43000)(9.8)(14)= \frac{1}{2}(5.9*10^5)X^2$

solving for x:

x = 4.4719 m