Question

An ore car of mass 39000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 25 m lower vertically, is a horizontally situated spring with constant 5.7 × 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 1

Answer:

x = 5.79 m

Explanation:

given,

mass of the car = 39000 Kg

spring constant = 5.7 x 10⁵ N/m

acceleration due to gravity = 9.8 m/s²

height of the track = 25 m

length of spring compressed = ?

using conservation of energy

potential energy is converted into spring energy

$m g h = \dfrac{1}{2}kx^2$

$x =\sqrt{\dfrac{2 m g h}{k}}$

$x =\sqrt{\dfrac{2\times 39000 \times 9.8 \times 25}{5.7 \times 10^{5}}}$

$x =\sqrt{33.5263}$

x = 5.79 m

the spring is compressed to x = 5.79 m to stop the car.