An elevator cab of mass m = 500 kg is descending with speed when its supposing cable begins to slip, allowing it to fall with constant acceleration.

(a) During ‘the fall through a distance d = 12 m, what is the work done on the cab by the gravitational force.

SOLUTION: The Key Idea here is that we can treat the cab as a particle and thus use the work Wt done on it by (a free body diagram of the cab with the cab’s displacement d included), we see that the angle between the directions of FR and the cab’s displacement.

(b) During the 12 m fall, what is the work done on the cab by the upward pull f of the elevator cable?

SOLUTION: A Key Idea here is that we can calculate the work with if we first find an expression for the magnitude T of the cable’s pull.

Next, substituting -for the (downward) acceleration a and then for the angle between the directions of forces f and mg, we find

(c) What is the net work W done on the cab during the fall?

SOLUTION: The Key Idea here is that the net work is the sum of the works done by the forces acting on the cab.

(d) What is the cab’s kinetic energy at the end of the 12 m fall?

SOLUTION: The Key Idea here is that the kinetic energy changes because of the-net work done on the cab, according.