Question

An elevator of mass M is at rest when its cable breaks. The elevator falls a distance h before it encounters a massless, cushioning spring (with spring constant k) at the bottom of the shaft. The spring has an uncompressed height of L. As the elevator falls, its safety clamp applies a frictional force of f on the elevator to help slow it down (this frictional force is much less than the force of gravity on the elevator)
1. Write an expression for the speed of the elevator before it encounters the spring.

Answer 1

Answer:$v=\sqrt{\dfrac{2[mgh+fh]}{m}}$

Explanation:

$\text{Given}$

cable breaks and elevator falls a distance h

Friction force f is applied by the clamps

Before it hits the spring velocity of elevator is given by the work energy theorem i.e. work done by all the forces is equal to change in kinetic energy            

$W_g+W_f=\text{change in kinetic energy}$

Where ,

$W_g=\text{Work done by gravity}$

$W_f=\text{Work done by friction}$

$mgh+f\cdot h=\frac{1}{2}mv^2$

$v=\sqrt{\dfrac{2[mgh+fh]}{m}}$