Aheavy rope, 50 ft long, weighs 0.5 lb/ft and hangs over the edge of a building 120 ft high. how much work is done (in ft-lb) in pulling half the rope to the top of the building?

The formula used in calculating for values of work is given as:

W = F d

However on the above formula, the force is constant but i this case, it is changing. Therefore we make use of the general form with the integral notation: 
W = ∫F(x) dx

with limits of x from 0 to 25 since only half of the rope is be pulled

The equation for F (x) is equal to the product of mass density and length x:
F(x) = 0.5x 
Substituting this into the integral work equation:
W = 0.5∫x dx

W = 0.5(0.5) [x2^2 – x1^2]                  ---> x from 0 to 25

W = 0.25 [25^2 -0]

unbalanced force

explanation:

we have been given a quadratic function and we need to restrict the domain such that it becomes a one to one function.

we know that vertex of this quadratic function occurs at (5,2).

further, we know that range of this function is .

if we restrict the domain of this function to either or , it will become one to one function.

let us know find its inverse.

upon interchanging x and y, we get:

let us now solve this function for y.

hence, the inverse function would be if we restrict the domain of original function to and the inverse function would be if we restrict the domain to .