In a shipping yard, a crane operator attaches a cable to a 1,390 kg shipping container and then uses the crane to lift the container vertically at a constant velocity for a distance of 33 m. Determine the amount of work done (in J) by each of the following. a) the tension in the cable.
b) the force of gravity.

a)  A = 449526  J,  b) 449526 J

Explanation:

In this exercise they do not ask for the work of different elements.

Note that as the box rises at constant speed, the sum of forces is chorus, therefore

           T-W = 0

           T = W

           T = m g

           T = 1,390 9.8

           T = 13622 N

Now that we have the strength we can use the definition of work

           W = F .d

            W = f d cos tea

a) In this case the tension is vertical and the movement is vertical, so the tension and displacement are parallel

              A = A  x

              A = 13622  33

               A = 449526  J

b) The work of the force of gravity, as the force acts in the opposite direction, the angle tea = 180

               W = T x cos 180

               W = - 13622 33

               W = - 449526 J

305.7

explanation:

density =m/v

density =917/3.00

density = 305.

approximately = 305.7

in four significant figures