Having trouble in my calculus 2 class with this question about Work done by a elevator motor

Question

An elevator is constructed using an electric motor at the top which is used to lift the elevator's car via a cable.

The cable weighs 6 pounds per foot. The cable is 100 feet long when the elevator car is at the first floor, and it is effectively 0 feet long when the car is at the top floor. The elevator's car weighs 500 pounds. How much work W does the motor do from lifting the cable and car when it takes the car from the first floor to the top floor?

I have been able to determine the shape of the solid and give its diameters. the problem I am having is how to calculate the 500 pounds into my integral equation.

Answer 1

You can compute the works done on the car and cable separately, without resorting to integration.

The work to lift the car to height $H$ is,

$$W_{car} = m_{car}gH = 500\times 32\times 100$$

where 32 is the gravitation constant in ft unit. And the work to lift the cable,

$$W_{cable} = m_{cable}g\frac{H}{2} = 600\times 32\times 50$$

Note that the height to be lifted is only half of the cable length because the center of mass for the cable is at its half way point. Alternatively, you could obtain it from integrating $6\cdot 32\int_0^{100} hdh $.

Then adding them together to obtain the total work.

Answer 2

Work is equal to the integral of the force over the displacment

At the first floor ( from which the elevator starts) the cable length is 100 and then as the elevator rises it goes to zero. Gravitational force points in the opposite direction that the displacement so we have

$$W=\int_{100}^{0}-g(6x+500 )dx $$ where x is the cable length