A boy lifts a load of 40 kgf through a vertical height of 2m in 5s by using a single fixed pulley when he applies an effort of 48 kgf.

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asked Dec 8, 2018 in Class X Science by priya12 (-12,494 points)

A boy lifts a load of 40 kgf through a vertical height of 2m in 5s by using a single fixed pulley when he applies an effort of 48 kgf. Calculate: (i) the mechanical advantage, and (ii) the efficiency of the pulley. Why is the efficiency of the pulley is not 100%? (iii) the energy gained by the load in 5s, and (iv) the power developed by the boy in raising the load.

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answered Dec 8, 2018 by navnit40 (-4,698 points)

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

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