Given: L = 40 kgf,
dL = 2 m,.
t = 5s,
E = 48 kgf.
(i) Mechanical advantage (M.A.) = L/E = 40/48 = 5/6 = 0.833
(ii) If the effort moves a distance d downwards, the load also moves a distance d upwards. So velocity ratio (V.R.) = d/d = 1,
Efficiency = M.A./V.R. = 0.833/1 = 0.833(or 83.3%).
The efficiency of the pulley is not 100% because some energy is wasted in overcoming the friction in the pulley bearings.
(iii) The energy gained by the load in 5s = Load x Displacement of load in 5s
= 40 kgf x 2 m = 80 kgf x m.
(iv) Power developed by the boy = Effort x Displacement of effort/Time = 48 kgf x 2m/5s
= 19.2 kgf x ms-1