Question

The mechanical advantage of a machine is 5 and its efficiency is 80%. It is used to lift a load of 200 kgf to a height of 20 m. Calculate:
(i) The effort required, and
(ii) The work done on the machine (g = 10 ms⁻²).

Answer 1

Given: M.A. = 5, Efficiency = 80% = 0.8

Load L = 200 kgf = 200 × 10 N = 2,000 N, d_L = 20 m.

(i) M.A. =`"L"/"E"`

∴ Effort, E = `"L"/"M.A."=200/5` kgf

= 40 kgf.

(ii) Work output = Load × Displacement of load 

= 2,000 N × 20 m 

= 40,000 J

Efficiency = `"Work output"/"Work input"`

∴ Work input = `"Work output"/"Efficiency"=40000/0.8`

= 50,000 J.