The efficiency of a machine is 60℅,find In joules the work done by a student using the machine to rise a load of 150kg through a vertical distance or 2.5 m (g=10m/s2)
Please help me to solve the problem
work input*.60=work output
work input(.60)=150kg*(9.8N/kg)*2.5m
work input= 150*9.8*2.5/.60
I would Never use g=10m/s^2, as it is not that value anywhere on Earth.
so the answer
physic
Thanks but i need the formular
Well, let's start by calculating the work done by the student using the machine. The formula for work is given by:
Work = Force x Distance x Cos(θ)
In this case, the force is equal to the weight of the load. So, let's calculate that first:
Weight = mass x gravity
Weight = 150kg x 10m/s²
Weight = 1500N
Now that we have the weight, we can calculate the work done:
Work = Force x Distance x Cos(θ)
Work = 1500N x 2.5m x Cos(θ)
However, we haven't been given the angle (θ) of the incline. So, we'll have to assume it's a vertical lift, and thus the angle is 0°. In this case, cos(0°) = 1.
Work = 1500N x 2.5m x Cos(0°)
Work = 1500N x 2.5m x 1
Finally, we can calculate the work done by the student:
Work = 3750 joules
So, the work done by the student using the machine to raise the load is 3750 joules. And let's hope they didn't clowns it up!
To find the work done by a student using the machine, we need to consider the efficiency of the machine. The efficiency of a machine is defined as the ratio of the useful work output to the total work input, expressed as a percentage.
Given:
Efficiency of the machine = 60% = 0.60
Load = 150 kg
Vertical distance = 2.5 m
Acceleration due to gravity (g) = 10 m/s^2
1. First, let's find the total work input.
The total work input is the product of the force applied (work input force) and the distance covered by the force. In this case, the force applied is equal to the weight of the load.
Weight of the load = mass × gravitational acceleration
Weight = 150 kg × 10 m/s^2 = 1500 kg⋅m/s^2 = 1500 N
Total work input = Force × Distance
Total work input = 1500 N × 2.5 m = 3750 N⋅m = 3750 J
2. Next, let's calculate the useful work output.
The useful work output is the work done against the force of gravity, which can be calculated using the formula:
Useful work output = force × distance
Useful work output = weight × distance
Useful work output = 1500 N × 2.5 m = 3750 N⋅m = 3750 J
3. Finally, let's find the work done by the student.
The work done by the student is equal to the useful work output divided by the efficiency of the machine, expressed as a decimal.
Work done by student = (Useful work output / Total work input) × 100%
Work done by student = (3750 J / 3750 J) × 0.60 = 0.60
Therefore, the work done by the student using the machine to lift a load of 150 kg through a vertical distance of 2.5 m is 0.60 Joules.