Question

A 500 kg elevator accelerates upward at 1.9 m/s2 for 20 m, starting from rest.

a) How much work does gravity do on the elevator?

b) How much work does the tension in the elevator cable do on the elevator?

c) What is the elevator’s kinetic energy after traveling 20 m?

Answer 1

(A)

The expression for the work done by the gravity, when the elevator moves up a distance is,

Here, is the work done by the gravity, is force due to gravity, and is the height.

The gravity acts downward and the displacement of the elevator is upward .

Substitute for and for .

= - 500*9.8*20 =

the work done by the gravity is =   -98000 J

B

Expression for the net force acting on the elevator is,

Rearrange the above expression for the magnitude of the tension.

  

= 500* ( 1.9+9.8) = 5850 N

The expression for the work done by the tension is,

Here, is the work done by the tension, is the tension in the elevator cable and is the distance travelled by the elevator.

The tension and the displacement are in same direction, so .

W = 5850*20*cos0 =117000 J  

The work done by the tension in the elevator cable is =117000 J

C)

The work energy theorem states that the net work done on an object is equal to the change in kinetic energy of the object.

Mathematically, it is expressed as,

Substitute for .

Here, is the initial kinetic energy and is the final kinetic energy.

The elevator starts from rest and hence the initial kinetic energy is zero.

KE = 117000 - 98000   =19000

The kinetic energy of the elevator, as reaches the 20 m height is = 19000 j