Answer to Question #102344 in Classical Mechanics for Udoy

Question

a lift starts falling due to its broken wire with a constant velocity of 8m/s . as safety measure the bottom has a 5m long spring on the ground and the lift stops 2 m above the ground by the spring.What is the spring constant of the spring?

Answer 1

The lift falls with the constant velocity, so we can say that the initial energy of the lift is

$$E_i = \frac{mv^2}{2}$$

When it collides with the spring, all his kinetic energy converts to the potential energy of the spring:

$$E_f = \frac{k\Delta x^2}{2}$$

where $\Delta x$ is the spring compression. As $E_i=E_f$, we can find the spring constant:

$$k = \frac{mv^2}{\Delta x^2}=\frac{8^2}{(5-2)^2}m=7.1m$$