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
To find the work in, work out, IMA, AMA, efficiency, and power in this scenario, we'll need to follow a step-by-step approach:
1. Work in: Work in represents the work done on the system. In this case, the student is pushing the cart up the ramp, so the work is done by the student against the force of gravity. The formula for work is given by W = F * d * cos(theta), where F is the force applied, d is the distance, and theta is the angle between the force and the displacement.
In this case, the force applied by the student is 70 N, the distance is the vertical height of the platform (1.8 m), and because the force is applied upwards against gravity, the angle theta is 0 degrees. So, work in = 70 N * 1.8 m * cos(0) = 126 J.
2. Work out: Work out represents the work done by the system. In this case, the work is done against the gravitational force to lift the cart and the boxes. The formula remains the same as work in, but the angle theta will be 180 degrees since the force is opposite to the displacement.
The force opposing the motion is the weight of the cart and the boxes. The weight can be calculated by multiplying the mass by the acceleration due to gravity, which is approximately 9.8 m/s^2. The total mass of the cart and the boxes is 6 kg + 2 * 12 kg = 30 kg. So, the weight is 30 kg * 9.8 m/s^2 = 294 N.
Using the formula for work again, we have work out = 294 N * 1.8 m * cos(180) = -294 J (negative sign indicates the opposite direction of displacement).
3. IMA (Ideal Mechanical Advantage): The IMA represents the mechanical advantage of an ideal machine, considering only the physical attributes of the machine and not any losses due to friction or other factors. In this case, the ramp acts as an inclined plane, and the IMA can be calculated using the formula IMA = length of the inclined plane / height of the inclined plane.
Given that the ramp is inclined at an angle of 15 degrees, we can use trigonometry to find the length of the inclined plane. The length is given by the vertical height divided by the sine of the angle, so length = 1.8 m / sin(15) = 6.95 m.
Similarly, the IMA is calculated by dividing the length by the height: IMA = 6.95 m / 1.8 m = 3.86.
4. AMA (Actual Mechanical Advantage): The AMA represents the mechanical advantage of the actual system, including factors like friction, imperfections, and energy losses. The AMA can be calculated as the ratio of the output force to the input force.
In this case, the output force is the weight being lifted (294 N), and the input force is the force applied by the student (70 N). So, AMA = 294 N / 70 N = 4.2.
5. Efficiency: Efficiency is the ratio of useful work output to work input, expressed as a percentage. It represents how effectively the system converts input energy into useful output energy.
The formula for efficiency is given by Efficiency = (work out / work in) * 100%. In this case, using the values we've already calculated, efficiency = (-294 J / 126 J) * 100% = -233%.
Note: The negative sign in the efficiency indicates that the work done by the system is more than the work done on the system. This means there are energy losses due to factors like friction.
6. Power: Power represents how quickly work is done or energy is transferred. The formula for power is given by Power = work / time.
In this case, since the work done by the student is the input work, we can use work in as the numerator. The time taken to do the work is given as 42 seconds.
So, power = 126 J / 42 s = 3 W (Watts).
To summarize, in this scenario:
- Work in = 126 J
- Work out = -294 J
- IMA = 3.86
- AMA = 4.2
- Efficiency = -233%
- Power = 3 W