Question

A boy uses a single fixed pulley to lift a load of 400 N through a vertical height of 5 m in 10 s. The
effort applied by him at the free end of the string is 480 N.
a) What is the mechanical advantage of the pulley ?
b) What will the energy gained by the load in 10 s ?
c) How much power was developed by the boy in the raising the load ?

Answer 1

Answer:

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

Explanation: