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Calculating Mechanical Energy and Applying Conservation of Energy, Study notes of Physics

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.

Typology: Study notes

2021/2022

Uploaded on 09/12/2022

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Systems
and energy
Objectives
Define a physical system.
Calculate the mechanical energy of a physical system.
Demonstrate and apply the law of conservation of energy.
Assessment
1.A system in physics is best described as . . .
A.a collection of related objects and interactions.
B.the masses and velocities of a collection of objects.
C.the masses and velocities of a collection of objects,
and the forces that act on them.
D.the objects within a volume that you are interested in.
Assessment
2.A large bird with a mass of 1.0 kg is flying at a height of 10 meters,
at a speed of 10 m/s. What is the mechanical energy of the bird?
Assessment
3.A 30 kg cart released from rest slides down a frictionless ramp
that drops 10 m from top to bottom.
a)What is the change in the system that allows you to apply
conservation of energy?
b)What are the states of the cart before and after the change?
c)What is the initial mechanical energy of the system?
d)Write down a statement of conservation of energy for the cart.
e)Which forms of energy are zero?
f)Solve the equation for the speed of the cart at the bottom and
calculate its value in m/s.
A. The energy remains constant
B.
C.
D. The potential energy is conserved.
Assessment
4.Which of these statements is not true for a closed system?
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Systems

and energy

Objectives

  • Define a physical system.
  • Calculate the mechanical energy of a physical system.
  • Demonstrate and apply the law of conservation of energy.

Assessment

  1. A system in physics is best described as... A. a collection of related objects and interactions. B. the masses and velocities of a collection of objects. C. the masses and velocities of a collection of objects, and the forces that act on them. D. the objects within a volume that you are interested in.

Assessment

  1. A large bird with a mass of 1.0 kg is flying at a height of 10 meters, at a speed of 10 m/s. What is the mechanical energy of the bird?

Assessment

  1. A 30 kg cart released from rest slides down a frictionless ramp that drops 10 m from top to bottom. a) What is the change in the system that allows you to apply conservation of energy? b) What are the states of the cart before and after the change? c) What is the initial mechanical energy of the system? d) Write down a statement of conservation of energy for the cart. e) Which forms of energy are zero? f) Solve the equation for the speed of the cart at the bottom and calculate its value in m/s. A. The energy remains constant B. C. D. The potential energy is conserved.

Assessment

  1. Which of these statements is not true for a closed system?

Physics terms

  • system
  • open system
  • closed system
  • mechanical energy
  • law of conservation of energy

Equations

Equations

For any closed system that undergoes a change, the total energy before the change is the same as the total energy after the change.

Big idea

Law of Conservation of Energy

Energy can never be created nor destroyed, only changed from one form to another.

Open and closed systems

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.

Open and closed systems

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.

Before the change

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?

After the change

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?

Look at the energies

Energy

Which terms are zero?

Energy

Which terms are zero?

Energy Let the initial height be zero. This makes the potential energy zero.

Which terms are zero?

Energy The speed is zero at the highest point. This makes the final kinetic energy zero.

Simplify the expression

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...

Simplify the expression

Answer

Calculate the height. Known values

Example solution

If the ball is thrown up at 10 m/s, how high does it rise?

Review the solution steps

Energy Identify the before and after states of your system. Write down the relevant forms of energy before and after. Before After

Review the solution steps

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

17 m/s

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.

Assessment

  1. A system in physics is best described as... A. a collection of related objects and interactions. B. the masses and velocities of a collection of objects. C. the masses and velocities of a collection of objects, and the forces that act on them. D. the objects within a volume that you are interested in.

Assessment

  1. A system in physics is best described as... A. a collection of related objects and interactions. B. the masses and velocities of a collection of objects. C. the masses and velocities of a collection of objects, and the forces that act on them. D. the objects within a volume that you are interested in.

Assessment

  1. A large bird with a mass of 1.0 kg is flying at a height of 10 meters, at a speed of 10 m/s. What is the mechanical energy of the bird?

Assessment

  1. A large bird with a mass of 1.0 kg is flying at a height of 10 meters, at a speed of 10 m/s. What is the mechanical energy of the bird? a) What is the change in the system that allows you to apply conservation of energy?

Assessment

  1. A 30 kg cart released from rest slides down a frictionless ramp that drops 10 m from top to bottom. a) What is the change in the system that allows you to apply conservation of energy? The cart moves down 10 meters. b) What are the states of the cart before and after the change?

Assessment

  1. A 30 kg cart released from rest slides down a frictionless ramp that drops 10 m from top to bottom.

Assessment

  1. Which of these statements is not true for a closed system? A. The energy remains constant B. C. D. The potential energy is conserved.