A MACHINE OF EFFICIENCY 60% IS USED TO RAISE A LOAD OF 150kg THROUGH A VERTICAL DISTANCE OF 2.5m.CALCULATE THE WORK DONE [IN JOULES]. I tried E=WORK OUTUT/WORK INPUT times 100% 60=150 times 10 times 2.5/work input times 100 60 times work input=3750 times 100 work input=375000/60 work done=6250J
Wo = F*d=M*g * d = 150*9.8 * 2.5
= 3675 J. = Work output or work done.
Input = 3675/0.6, but it did not ask for the input.
Yes
To calculate the work done, you can use the formula:
Work Done = Work Input = (Efficiency/100) * (Load * Distance)
In this case, the efficiency is given as 60% (or 0.6), the load is 150 kg, and the distance is 2.5 m.
Work Done = (0.6) * (150 kg * 2.5 m)
= 0.6 * 375 kg*m
= 225 kg*m
Since 1 Joule is equal to 1 kg*m^2/s^2, we can convert the work done into Joules:
Work Done (in Joules) = 225 kg*m / 1
= 225 J
Therefore, the work done to raise the load of 150 kg through a vertical distance of 2.5 m is 225 Joules.
To calculate the work done using a machine with an efficiency of 60%, we can follow these steps:
1. Convert efficiency to decimal form: Efficiency is given as 60%. To convert it to decimal form, divide it by 100. Thus, the efficiency becomes 0.60.
2. Calculate the work done by the machine: The work output of the machine can be determined using the formula:
Work output = Efficiency * Work input
In this case, the work output is the work done by the machine, which we need to calculate. The efficiency is 0.60.
Work output = 0.60 * Work input
3. Substitute known values: The load being raised is 150 kg, and it is being lifted through a vertical distance of 2.5 m.
Work output = 0.60 * Work input
Work output = 150 kg * 9.8 m/s^2 * 2.5 m
Note: The acceleration due to gravity is approximately 9.8 m/s^2.
4. Solve for work input: Rearrange the equation to solve for Work input.
Work input = Work output / 0.60
Work input = (150 kg * 9.8 m/s^2 * 2.5 m) / 0.60
Work input = 6250 Joules
Therefore, the work done is 6250 Joules.