Under less than ideal conditions, a student pushes a 6 kg cart holding 2 boxes of books up a ramp to a platform that is 1.8 m high in 42 sec. Each box contains 12 kg of books. The ramp is inclined at an angle of 15 degrees. He pushes w/ a steady force of 70 N to move it up.
Find: -Work in
-Work out
-IMA
-AMA
-Efficiency
-Power
This is relatively simple, I will be happy to critique your thinking.
To find the work in, work out, IMA (Ideal Mechanical Advantage), AMA (Actual Mechanical Advantage), efficiency, and power, we'll go step by step.
1. Work in:
The work done on an object is calculated using the formula: Work = Force × Distance × Cos(angle)
In this case, the force applied by the student is 70 N, and the distance is the vertical height the cart is lifted, which is 1.8 m. The angle is not relevant for this calculation. Therefore, the work in can be calculated as:
Work in = 70 N × 1.8 m × Cos(0°) = 70 N × 1.8 m = 126 Joules.
2. Work out:
The work done by the student to move the cart against gravity is equal to the work done against the weight of the cart plus the work done against the weight of the boxes. The work done against gravity is equal to the total weight of the cart and boxes multiplied by the vertical height they were lifted.
Total weight = (Mass of cart + Mass of boxes) × Gravitational acceleration
The gravitational acceleration is approximately 9.8 m/s^2.
Total weight = (6 kg + 2 kg) × 9.8 m/s^2 = 64.4 N
Work out = Total weight × Distance × Cos(angle) = 64.4 N × 1.8 m × Cos(15°) ≈ 171.64 Joules.
3. IMA (Ideal Mechanical Advantage):
The IMA can be calculated using the formula:
IMA = (Length of slope or ramp)/(Height of slope or ramp)
In this case, the angle of inclination is given, which is 15°. Length of the slope is not provided. Therefore, without the length of the slope, we cannot accurately calculate the IMA.
4. AMA (Actual Mechanical Advantage):
AMA is the ratio of the output force exerted by the machine to the input force applied to the machine.
AMA = (Output Force)/(Input Force)
In this case, the force applied by the student is the input force, which is 70 N. The weight of the cart and boxes is the output force.
Total weight = (Mass of cart + Mass of boxes) × Gravitational acceleration = (6 kg + 2 kg) × 9.8 m/s^2 = 64.4 N
AMA = Total weight / Input Force = 64.4 N / 70 N = 0.92
5. Efficiency:
Efficiency is the ratio of the useful work output to the total work input. It is often expressed as a percentage.
Efficiency = (Work out / Work in) × 100
Efficiency = (171.64 Joules / 126 Joules) × 100 = 136.19%
6. Power:
Power is the rate at which work is done or the work done per unit time.
Power = Work / Time
In this case, the time taken is given as 42 seconds.
Power = Work out / Time = 171.64 Joules / 42 s ≈ 4.087 Watts
Therefore, the answers are:
- Work in = 126 Joules
- Work out ≈ 171.64 Joules
- IMA: Cannot be calculated without the length of the slope.
- AMA = 0.92
- Efficiency = 136.19%
- Power ≈ 4.087 Watts