# Thread: calc find the work done

1. ## calc find the work done

A 40 ft long rope weighs 0.35 lb/ft and hangs over the edge of a building 100 ft high. How much work is done in pulling the rope to the top of the building

2. Originally Posted by harry
A 40 ft long rope weighs 0.35 lb/ft and hangs over the edge of a building 100 ft high. How much work is done in pulling the rope to the top of the building
There is a short cut for this, as the work is the same as that required to lift a
point mass equal to that of the rope from a position coincident with the
centre of mass of the rope to the top of the building. This is also equal to
the change in potential energy of the system so is:

W = mgh,

here m = 0.35*40 = 14 pounds, g= 32 ft/s^2, and h = 20 ft, so:

W = 8960 ft poundals (or 280 ft pounds)

RonL

3. Originally Posted by harry
A 40 ft long rope weighs 0.35 lb/ft and hangs over the edge of a building 100 ft high. How much work is done in pulling the rope to the top of the building
i guess you may also want to know how to do it the integral way.

since we want to take the integral, we begin by dividing the rope into small pieces of length delta x. since the rope weighs 0.35 lb/ft, the weight of each piece is 0.35*delta x. we are moving the rope a distance x, we call the distance x and not 40 since it is always changing (see diagram). so now we have the force ( 0.35*delta x) and the distance ( x ).

now work, by the riemann sum is given by:

W = lim{n-->infinity}E 0.35 * deltax*x = lim{n-->infinity}E 0.35 x deltax
please note that E is the summation sign. changing this to an integral we get:

W = int{0.35x}dx ............the delta x becomes the dx

you dont have to go through all of this everytime you do a problem. just remember work = int{force*distance}dx, call the distance x (usually the case) and the force is the weight per unit length.

so W = int{0.35x}dx evaluated between 0 and 40
=> W = [0.35/2 x^2]
=> W = [0.35/2 * 40^2 - 0.35/2 * 0^2]
=> W = 280 - 0 = 280 ft-lb