Question

An elevator of mass 2 metric tons is moving downward with speed 1.5 m s^-1. Suddenly the breaking system that keeps in moving at the above speed fails and the elevator free falls for 5 m, before hitting a safety spring of constant 1.5 104N m^-1. Find by how much the spring will compress. Answer in meters [10 points].

Answer 1

Free fall means, acceleration a = g = 9.81 m/s^2

so, speed of the elevator after 5 m fall -

v^2 = u^2 + 2*a*s

=> v^2 = 1.5^2 + 2*9.81*5

=> v^2 = 2.25 + 98.1 = 100.35

=> v = 10.02 m/s

now, suppose the spring compresses to a distance of x meter.

apply conservation of energy -

kinetic energy of the lift = spring energy of the spring

=> (1/2)*m*v^2 = (1/2)*k*x^2

=> 0.5*2000*10.02^2 = 0.5*1.5 x 10^5 * x^2

=> x^2 = (2000*10.02^2) / (1.5 x 10^5) = 1.339

=> x = 1.157 m