Question

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.

Answer 1

a) A = 449526 J, b) 449526 J

Step-by-step 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