Question

An electric elevator with a motor at the top has a multistrand cable weighing 7 lb divided by ft. When the car is at the first​ floor, 160 ft of cable are paid​ out, and effectively 0 ft are out when the car is at the top floor. How much work does the motor do just lifting the cable when it takes the car from the first floor to the​ top?

Answer 1

Answer:

The amount of work is 89600 ft-lb.

Explanation:

Given that,

Length of the cable = 160 ft

Weight density = 7 lb/ft

Weight of cable to be filled  = 7(160-x)

We need to calculate the amount of work

Using formula of work done

$\int{dW}=\int_{0}^{160}{7(160-x)}dx$

On integration

$W=[1120x-\dfrac{7x^2}{2}]_{0}^{160}$

$W=1120\times160-0-\dfrac{7\times(160)^2}{2}+0$

$W=89600\ ft-lb$

Hence, The amount of work is 89600 ft-lb.

Answer 2

Answer:

The amount of work is 89600 ft-lb.

Explanation:

Given that:

Length of the cable = 160 ft

Weight density = 7 lb/ft

Weight of cable to be filled  = 7(160-y)

We need to calculate the amount of work

Using formula of work done:

$W = \int\limits^a_b {7*(160-y)} \, dy \\\\W = {7*(160y-y^2/2)}\\a = 160\\b = 0\\\\W = 7*(160^2-160^2/2)\\W = 89,600ft-lb$

Hence, The amount of work is 89,600 ft-lb.