A fixed pulley is driven by a 100 kg mass falling at a rate of 8.0 m in 4.0 s. It lifts a load of 75.0 kgf.

Calculate —

(a) The power input to the pulley taking the force of gravity on 1 kg as 10 N.

(b) the efficiency of the pulley, and

(c) the height to which the load is raised in 4.0 s.

(a) As we know,

$\text {Power} = \text {Force} \times \text {velocity} \\[0.5em]$

and

$\text {Velocity} = \dfrac{\text{Change in distance}}{\text{Change in time}} \\[0.5em] \text {velocity} = \dfrac{8}{4} \\[0.5em] So, \text {velocity} = 2 m/sec \\[0.5em]$

Given,

$\text {Tension in pulley} = 100 \times 10 \\[0.5em] \text {Force} = \text {Tension} = 1000 N \\[0.5em]$

Substituting the values in the formula for power we get,

$\text {Power} = 1000 \times 2 \\[0.5em] \Rightarrow \text {Power} = 2000 W \\[0.5em]$

(b) We know,

$\eta = \dfrac{M.A.}{V.R.} \\[0.5em]$

Given,

$\text {Load} = 75 \times 10 \\[0.5em] \text {Load} = 750 N \\[0.5em]$

We know,

$M.A. = \dfrac{Load}{Effort}$ Substituting the values for M.A. we get,

$M.A. = \dfrac{750}{1000} \\[0.5em] \Rightarrow M.A. = 0.75 \\[0.5em]$

When the effort moves by a distance d downwards, the load moves by the same distance upwards. So we get,

V.R. = 1

Substituting the values of M.A. and V.R. in the formula for efficiency we get,

$\eta = \dfrac{0.75}{1} \\[0.5em] \Rightarrow \eta = 0.75 \\[0.5em]$

Hence, efficiency = 75 %

(c) As the load moves same distance upwards as much as effort moves downwards. So, the height achieved by the load is 8 m.