A machine of efficiency 80% is used to raise a load of body of mass 75kg through a vertical height of 3m in 30seconds.Calculate the power input.
E=Work done/txinput x100
80=75*10*3*100/30input
Cross multiply
80x30input =225000
2400input =225000
Input =225000/2400
Input =93.75
P = W/t = mgh/t = 75*9.8*3/30
80% eff means divide above by 0.8
Yes
Yes that's the question
Well, let's calculate the power input. To do this, we can use the formula:
Power = (Work Done) / (time)
First, we need to calculate the work done. The work done is equal to the force applied (mg) multiplied by the distance moved (height).
Work Done = mg * height
where m is mass and g is the acceleration due to gravity.
Work Done = 75 kg * 9.8 m/s^2 * 3 m
Work Done = 2205 Joules
Now, we can calculate the power input:
Power = Work Done / time
Power = 2205 J / 30 s
Power = 73.5 Watts
So, the power input is approximately 73.5 Watts. However, with 80% efficiency, we would multiply this by 0.8 to get the actual power input.
Actual Power Input = 73.5 W * 0.8
Actual Power Input = 58.8 Watts
So, the actual power input is approximately 58.8 Watts. That's the power of efficiency clowning around!
To calculate the power input, we need to use the formula:
Power = Work done / Time taken
First, let's find the work done.
Work Done = Force x Distance
The force required to lift the load is equal to the weight of the object, which can be calculated using the formula:
Force = Mass x Acceleration due to Gravity
= Mass x 9.8 m/s^2 (standard value for acceleration due to gravity)
So, the force required is:
Force = 75 kg x 9.8 m/s^2
= 735 N
Now, let's calculate the distance:
Distance = 3 m
Next, we need to calculate the work done:
Work Done = Force x Distance
= 735 N x 3 m
= 2205 Nm (Joules)
The time taken is given as 30 seconds.
Now, we can calculate the power input:
Power = Work Done / Time taken
= 2205 Nm / 30 s
= 73.5 W
Therefore, the power input required is 73.5 Watts.