Previously, you learned how to calculate the Interactive popup. Assistance may be required. mechanical energy of a system. In this lesson, you will concentrate more on how to use mechanical energy to make predictions about how the system will behave.

Mechanical Energy:
The sum of the kinetic energy and all forms of potential energy—measured in Joules (J)

Kinetic Energy:
The energy of motion → KE = ½ mv²

Potential Energy:
Stored energy

Gravitational: Energy stored due to height → PE_g = mgh
Elastic: Energy stored due to compression or stretching → PEs = ½ kx²

Once you know the total mechanical energy of a system, there are two possibilities. Either work is done on the system and the total energy changes, or no work is done and the system obeys the Interactive popup. Assistance may be required. the law of conservation of energy.

The Law of Conservation of Energy
In a closed isolated system, total mechanical energy is conserved

System:  Group of related objects
Closed:  No mass enters or leaves the system
Isolated:  No unbalanced outside forces—no work done on the system
Conserved:  Stays the same

KE_i + PE_gi + PE_si = KE_f + PE_gf + PE_sf

In the equation, the subscript i means initial and the subscript f means final.

In the following simulation:

1. Set the friction slider to ‘none’
2. Select ‘Show Energy of 1’
3. Select ‘1 over 2 1 16 time’
4. Use your mouse to add one of the 100 gram masses to the end of spring one
5. Stretch the spring so that the mass is almost at the bottom of the screen and then release the mass and watch the values on the graph

Masses and Springs

What happens to the kinetic energy as the mass moves up and down?

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Check Your Answer

It is zero at the top and bottom of the mass’s path and a maximum in the middle.

What happens to the gravitational potential energy as the mass moves up and down?

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Check Your Answer

It reaches a maximum at the mass’s highest point and a minimum at the mass’s lowest point.

What happens to the spring potential energy as the mass moves up and down?

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Check Your Answer

It is a maximum at the top and bottom of the mass’s path and zero in the middle.

What happens to the total energy as the mass moves up and down?

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Check Your Answer

It stays the same.

Calculating the Law of Conservation of Energy

Before beginning calculations, watch an example in the video below of how to solve a law of conservation of energy. In the video, you will see a few different examples. Pause and take notes on what you see.

Remember, sometimes when calculating information, you have to calculate one type of equation before calculating the final equation. This is called a two-step equation.

Below are equations that may require more than one step. Use equations found on the Reference Chart when attempting example problems for the following:

Kinetic Energy equation

Potential energy – gravitational equation

Potential energy elastic – equation, and velocity

Example Problem 1

A 45 kg child is playing on a swing. At her highest point the swing is 1.75 meters above the ground, and at the lowest point it is 0.45 meters above the ground. How fast is she going at the lowest point?

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Hint 1

Conservation of Energy says: KE_i + PE_gi + PE_si = KE_f + PE_gf + PE_sf

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Hint 2

There are no springs in this problem, so you can drop PE_s out of both sides of the equation:
KE_i + PE_gi= KE_f + PE_gf

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Hint 3

At her highest point, her velocity is zero.

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Solution

Given: m = 45 kg    h₁ = 1.75 m    v₁ = 0 m/s    h₂ = 0.45 m
Unknown: v₂ = ?
Equation: KE_i + PE_gi = KE_f + PE_gf
½ mv₁² + mgh₁ = ½ mv₂² + mgh₂
Simplify:

v₁ = 0, so the equation becomes: mgh₁ = ½ mv₂² + mgh₂
The mass cancels from both sides of the equation, so it becomes: gh₁ = ½ v₂² + gh₂

Solve:

gh₁ = ½ v₂² + gh₂
(9.8) (1.75) = ½ v₂² + (9.8) (0.45)
17.15 = ½ v₂² + 4.41
12.74 = ½ v₂²
v₂² = 25.48
v₂ = 5.05 m/s

Example Problem 2

A 2 kg mass is sliding on a table at 3 m/s. It hits the end of a long relaxed spring with spring constant k = 100 N/m. How far will the mass compress the spring before coming to rest?

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Hint 1

Conservation of Energy says: KE_i + PE_gi + PE_si = KE_f + PE_gf + PE_sf

Interactive popup. Assistance may be required.

Hint 2

There is no change in height in this problem, so PE_g drops from both sides of the equation:
KE_i + PE_si= KE_f + PE_sf

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Hint 3

There are 2 hidden zeros in the problem:
Relaxes spring→x_i = 0
Coming to rest→v₂ = 0

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Solution

Given: m = 2 kg    v₁ = 3 m/s   k = 100 N/m    x₁ = 0m    v₂ = 0 m/s
Unknown: x₂ = ?
Equation: KE_i + PE_si = KE_f + PE_sf
½ mv₁² + ½ kx₁² = ½ mv₂² + kx₂²

Simplify:

x₁ and v₂ are both zero, so the equation becomes: ½ mv₁² = kx₂²
If we multiply both sides by 2, the equation becomes: mv₁² = kx₂²

Solve:

mv₁² = kx₂²
(2)(3²) = 100 x²₂
0.18 = x²₂
x₂ = 0.42 m