A machine with velocity ratio of 5 was used to raise a load of 1.2kg mass through a height of 10m .If the work done to raise the load is 800J . Find the useful work output , efficiency of the machine
The work done was mgh or 1.2*9.8*10=118 joules
efficiency=118/800
To find the useful work output and efficiency of the machine, we can use the following formulas:
Useful work output = Work input – Work done against friction
Efficiency = (Useful work output / Work input) × 100%
Given:
Velocity ratio (VR) = 5
Mass of the load (m) = 1.2kg
Height (h) = 10m
Work done (W) = 800J
Let's calculate the values step by step:
Step 1: Calculate the work input
Work input (Wi) = m × g × h
Where g is the acceleration due to gravity (approximately 9.8 m/s²)
Wi = 1.2kg × 9.8 m/s² × 10m
Wi = 117.6 J
Step 2: Calculate the useful work output
Since there is no mention of friction, we assume there is no work done against friction.
Useful work output (Wu) = Work input
Wu = 117.6 J
Step 3: Calculate the efficiency
Efficiency (η) = (Useful work output / Work input) × 100%
η = (117.6 J / 117.6 J) × 100%
η = 100%
Therefore, the useful work output is 117.6 J, and the efficiency of the machine is 100%.
To find the useful work output and efficiency of the machine, we need to use the following formulas:
1. Useful work output = Input work - Work lost due to friction or other factors
2. Efficiency = (Useful work output / Input work) x 100%
First, let's find the input work:
Input work = force applied x distance
Force applied = mass x acceleration due to gravity (g)
Given:
- Mass of the load = 1.2 kg
- Height = 10 m
- Velocity ratio = 5
To find the force applied, we can use the formula:
Force applied = mass x g
Where g is the acceleration due to gravity (approximately 9.8 m/s^2).
Force applied = 1.2 kg x 9.8 m/s^2 = 11.76 N
To find the distance, we can use the formula:
Distance = height / velocity ratio
Distance = 10 m / 5 = 2 m
Now, we can calculate the input work:
Input work = Force applied x Distance
Input work = 11.76 N x 2 m = 23.52 J
So, the input work is 23.52 J.
Next, let's find the useful work output:
Useful work output = Input work - Work lost
Given that the work done to raise the load is 800 J, we can calculate the work lost:
Work lost = Input work - Useful work output
800 J = 23.52 J - Useful work output
Useful work output = 23.52 J - 800 J
Useful work output = -776.48 J (negative because it is lost work)
The useful work output is -776.48 J (or approximately -776.5 J).
Finally, let's calculate the efficiency:
Efficiency = (Useful work output / Input work) x 100%
Efficiency = (-776.48 J / 23.52 J) x 100%
Efficiency = -33% (approximately)
Therefore, the useful work output is approximately -776.5 J and the efficiency of the machine is approximately -33%.