e = L/EVr = 50*9.8/(110*7) = 0.64 = efficiency.
a. W = L*d = (50*9.8)*0.8 = 392 J. = Work done.
e = 392/110d= 0.64
70.4d = 392
d = 5.56 m.
b. 110*5.56 = ___ Joules.
c. (110*5.56)-392 = ___ Joules.
A block and tackle with a velocity ratio of 7 is used to raise a mass of 50kg through a vertical distance of 800mm at a steady rate. If the effort is equal to 110N. Determine
(a) The distance moved by the effort.
(b) The workdone by the effort in lifting the load.
(c) The loss in energy involved in operating the machine.
1 answer