The tower crane will lift a 200 kg load with an acceleration of 0.1 m / s square for 10 seconds. what work did he do?

Initial data: t (duration of the tower crane) = 10 s; m (mass of the lifted load) = 200 kg; a (acceleration with which the tower crane lifted the load) = 0.1 m / s2; V0 (initial speed of the load) = 0 m / s.

1) Determine the speed that the load will develop at the end of the ascent: V = V0 + a * t = a * t = 0.1 * 10 = 1 m / s.

2) Calculate the work that the tower crane did: A = ΔEk = (m * V ^ 2/2) – (m * V0 ^ 2/2) = m * V ^ 2/2 = 200 * 12/2 = 100 J …

Answer: The tower crane will do 100 Joules.