the efficiency of a machine is 80% determine the work done of a person using the machine to raise a load of 250kg through a vertical distance of 3.0m (take g =10ms-2).
g is not 10m/s^2 anywhere on Earth. I have never understood folks doing that, unless they do not know how to use a calulator.
work= 250*g*3.0 /.80
F = M*g = 250*10 = 2500 N.
Wo = F*d = 2500*2500*3 = 7500 J.
Wo/Wi = 0.8,
7500/Wi = 0.8,
Wi =
To determine the work done by a person using the machine, we first need to understand the concept of efficiency. Efficiency is the ratio of useful work output to the total work input. In this case, the efficiency of the machine is given as 80%, which means that 80% of the work input is converted to useful work output.
To calculate the work done by a person using the machine, we need to know the work input, which is the work done against gravity to lift the load, and the work output, which is the useful work done.
First, let's calculate the work input:
Work input (W_in) = force × distance
The force required to lift the load is equal to the weight of the load:
force = mass × acceleration due to gravity
= 250 kg × 10 m/s^2
= 2500 N
Distance = 3.0 m
Work input (W_in) = 2500 N × 3.0 m
= 7500 N·m (Joules)
Now, let's calculate the work output using the efficiency:
Efficiency (η) = (Work output / Work input) × 100%
Rearranging the equation, we can calculate the work output:
Work output = Efficiency × Work input
Substituting the given values:
Work output = 0.8 × 7500 N·m
= 6000 N·m (Joules)
Therefore, the work done by the person using the machine to raise the load is 6000 Joules.