Question

the end of the tracks, 8.8 m lower vertically, is a horizontally situated spring with constant 5 × 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

Assuming the mass of the car, m = 43000 kg

initial speed u = 0

vertical distance moved, h = 8.8 m

spring constant k = 5 x  10⁵ N / m

acceleration of gravity = 9.8 m/s²

From law of conservation of energy ,

Gravitational potential energy at starting position =potential energy of the spring at maximum compression

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

                $x = \sqrt{\dfrac{2mgh}{k}}$

                $x = \sqrt{\dfrac{2\times 43000\times 9.8\times 8.8}{5\times 10^5}}$

                    x = 14.83 m

If the mass of the car is equal to 43000 Kg the spring is compressed to 14.83 m