Question

At a dock, a crane lifts a 2009 kg container 20 m, swings it out over the deck of a freighter, and lowers the container into the hold of the freighter, which is 8 m below the level of the dock. The acceleration of gravity is 9.81 m/s 2 . Neglect friction losses. How much work is done by the crane on the container? Answer in units of kJ

Answer 1

Answer:

W = 157.5kJ

Explanation:

Assuming it moves the container at constant speed, the work done by the crane will be equal to the variation of the potential gratitational energy on the container:

$Wc = \Delta E = m*g*(h2 - h1)$  where h2= -8m  and  h1=0m

Wc = 157.5kJ