a/ Calculate the lift workb/ What is the power used by the crane when the efficiency is 60%? (Can you explain for me what efficiency means)c/ What will the kinetic energy of the load be when it hits the ground, if it falls from the height 8.5 meters, g=9.81 m/s^2Thank you.
The work done (W)= Force*Distance. Force is 9.81*3000kg. So W = 9.81*3000*8.5 = 250,155 J Power (P)= Work(W)/Time(t) = 250,155/15 = 16,677 W 60% efficiency means that the lift is only using 60% of its maximum amount of power output. So to find 100% (lets call it x) of the crane's efficiency, consider the following: x * 0.6 = 16,677 x = 16,677 / 0.6 = 27,795 W Total mechanical energy in a system remains constant, so the kinetic energy the moment it hits the ground equals the potential energy when it is still hanging in the air. Potential energy = mgh = 3000x9.81x8.5 = 250,155 J You know this is correct, because it is the same amount of work the crane performed on the object to lift it.
evaluate the equation P = W/t the place W = F x d (F = tension (load) d = distance moved interior the path of tension, W = artwork carried out/ability transferred, P = ability, t = time) a million) if P = W/t then t = W/P yet W = F x d so t = F x d/P F = 6N d = 2m P = 3W t = 6 x 2/ 3 t = 4s 2) returned P = W/t = F x d/t so d = Pt/F d = 12W x 5s/ 6N d = 10m
I wonder if you saw my response to the additional question you asked of my previous answer to this question .. To see my response .. go to your previous question and click on comment in the Action Bar (below 'Asker's Rating')
