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

x = 5.79 m

Step-by-step 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 = (1)/(2)kx^2`

`x =\sqrt{(2 m g h)/(k)}`

`x =\sqrt{(2* 39000 * 9.8 * 25)/(5.7 * 10^(5))}`

`x =√(33.5263)`

x = 5.79 m

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