Question

A fixed pulley is driven by a 100 kg mass falling at a rate of $ 8.0m $ $ 4.0s $ . It lifts a load of $ 75.0kgf $ . Calculate,
A. the power input to the pulley taking the force of the gravity on 1kg as 10 N.
B. the efficiency of the machine, and
C. the height to which the load is raised in $ 4.0s $.

Answer 1

Hint: The velocity ratio of a single fixed pulley using an inextensible string is equal to 1. ecall that efficiency can be given as power output over power input.

Formula used: In this solution we will be using the following formulae;
 $ VR = \dfrac{{{d_e}}}{{{d_l}}} $ where $ VR $ is the velocity ratio of a simple machine, $ {d_e} $ is the distance moved by effort in a time, and $ {d_l} $ is the distance moved by load in the same time.
 $ Eff = \dfrac{{{P_{out}}}}{{{P_{in}}}} $ where $ Eff $ is the efficiency of the machine, $ {P_{out}} $ is the power output and $ {P_{in}} $ is the power input.
 $ P = Fv $ where $ P $ is power delivered by a force $ F $ and $ v $ is the velocity of the object it was delivered to.
 $ W = mg $ where $ W $ is the weight on a body, $ m $ is mass of a body, and $ g $ is the acceleration due to gravity.

Complete step by step answer:
For power we note that the input force is the 100 kg mass. Hence, we have the force is given by
 $ F = mg = 100 \times 10 = 1000N $
The mass moved at a rate of 8 m in 4 seconds, hence
 $ v = \dfrac{d}{t} = \dfrac{8}{4} = 2m/s $
Hence, power delivered can be given by
 $ P = Fv $ where $ P $ is power delivered by a force $ F $ and $ v $ is the velocity of the object it was delivered to. Hence,
 $ {P_{in}} = 1000 \times 2 = 2000{\text{W = 2kW}} $
For efficiency, we calculate the power output as
 $ {F_{out}} = 75kgf \times 10\dfrac{N}{{kgf}} = 750N $
Hence, the power output is
 $ {P_{out}} = 750 \times 2 = 1500{\text{W}} $ (note that the velocity input is equal to velocity output)
Efficiency can be given as
 $ Eff = \dfrac{{{P_{out}}}}{{{P_{in}}}} $ where $ Eff $ is the efficiency of the machine, $ {P_{out}} $ is the power output and $ {P_{in}} $ is the power input. Hence,
 $ Eff = \dfrac{{1500}}{{2000}} = 0.75 = 75\% $
Now, the distance in which the load is raised in 4 seconds will be calculated from velocity ratio, which is
 $ VR = \dfrac{{{d_e}}}{{{d_l}}} $ where $ VR $ is the velocity ratio of a simple machine, $ {d_e} $ is the distance moved by effort in a time, and $ {d_l} $ is the distance moved by load in the same time.
Velocity ratio of a single fixed pulley is equal to 1, hence
 $ 1 = \dfrac{8}{{{d_l}}} $
 $ \Rightarrow {d_l} = 8m $ .

Note:
For clarity, the velocity ratio of a single fixed pulley is equal to one because the velocity on both load side and effort side are equal. To understand why this is, you can follow the rationale that when force is applied at one end which moves the string by a particular amount, the same occurs at the other end as long as the string is inextensible.