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A problem-solving approach to calculating the mechanical energy of a system and applying the law of conservation of energy. It includes examples of a large bird in flight and a cart sliding down a ramp. Students will learn how to identify the relevant forms of energy and determine their values before and after a change in the system.
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For any closed system that undergoes a change, the total energy before the change is the same as the total energy after the change.
Energy can never be created nor destroyed, only changed from one form to another.
In an open system , energy and matter can pass through the imaginary system boundary and leave the system. A system is a group of interacting objects and influences, such as forces.
In an closed system , no energy and matter can pass through the system boundary. The energy in a closed system cannot change. A system is a group of interacting objects and influences, such as forces.
The ball leaves your hand with speed v. Let this represent the system before the change. Consider tossing a baseball straight up in the air. How high will it go?
The ball keeps rising, and slowing down, until it has zero speed at its maximum height. Choose this to be the system after the change. Consider tossing a baseball straight up in the air. How high will it go?
Energy
Energy
Energy Let the initial height be zero. This makes the potential energy zero.
Energy The speed is zero at the highest point. This makes the final kinetic energy zero.
Energy We are left with this statement of the conservation of energy for the ball. Divide by m on both sides and solve for h...
Calculate the height. Known values
If the ball is thrown up at 10 m/s, how high does it rise?
Energy Identify the before and after states of your system. Write down the relevant forms of energy before and after. Before After
Energy Conservation of energy problems might include elastic potential energy —in addition to gravitational potential and kinetic energy. Before After
Complete the tables by calculating the energies. Start with the potential energies.
m = 2,000 kg hi = 30 m hf = 16 m What is the total mechanical energy at Point 1? at Point 2?
m = 2,000 kg hi = 30 m hf = 16 m What is the final kinetic energy?
m = 2,000 kg hi = 30 m hf = 16 m How did you determine the final kinetic energy?
m = 2,000 kg hi = 30 m hf = 16 m What is the final speed of the cart?
m = 2,000 kg hi = 30 m hf = 16 m
m = 2,000 kg hi = 30 m hf = 16 m
Kinetic energy depends on speed and mass. If you know the kinetic energy and the mass you can always calculate the speed.
m = 2,000 kg hi = 30 m hf = 16 m There is another way to get the solution: solve for vf algebraically. First, eliminate terms that are zero.
m = 2,000 kg hi = 30 m hf = 16 m 1 2 m = 2,000 kg hi = 30 m hf = 16 m What is the next step? Divide by m and multiply by 2.
m = 2,000 kg hi = 30 m hf = 16 m What is the next step? Group the terms containing like variables.
m = 2,000 kg hi = 30 m hf = 16 m What’s next? Take the square root of both sides to get the desired result.
m = 2,000 kg hi = 30 m hf = 16 m
If you know the potential energy (and mass), then you can calculate the height.
m = 2,000 kg hi = 30 m vf = 20 m/s 188, 400, 588, If you know the potential energy (and mass), then you can calculate the height. The final height is 9.6 meters
m = 2,000 kg hi = 30 m vf = 20 m/s 188, 400, 588, Let’s go through the algebraic solution. Solve this equation for hf in terms of hi and vf. The initial kinetic energy is zero. What’s next?
m = 2,000 kg hi = 30 m vf = 20 m/s
m = 2,000 kg hi = 30 m vf = 20 m/s Cancel out the masses. Get the term containing hf alone on one side. 1 2 m = 2,000 kg hi = 30 m vf = 20 m/s Here is the equation for hf in terms of hi and vf. Solve for hf.
m = 2,000 kg hi = 30 m vf = 20 m/s Here is the equation for hf in terms of hi and vf. Solve for hf.