Conservation of Energy

Take a few minutes to do the Energy Conservation Inquiry

What conclusions can be made?

Without friction

With friction

ME stayed the same

KE and PE traded off but always added up to ME

PE increased with greater heights

KE increased when PE decreased

ME reduced

Everything else did the same thing as without friction

The cart would eventually come to a full stop

Energy Bar Graphs

• Show relative amounts of KE, PE, and ME in a system
• Sometimes they can show a maximum ME value
• Put an ‘X’ in a column that has zero value

Law of Conservation of Energy

Energy cannot be created or destroyed.

Work can transform energy from one form to another, but the total amount of energy never changes.

Examples of Energy Transformations:

Electric → Light + Heat

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Elastic PE → KE

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Gravitational PE → KE

Energy Transformations in Toys

With your group try to figure out ALL the energy transformations that are happening in your toy.

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Pick someone to present to the class.

Conservation of Energy

Ignoring friction, total mechanical energy is conserved:

In other words, the total amount of energy stays the same, it just transforms into different types of energy

ME Conservation Example: Frictionless Skiing

What do you notice about the KE and PE at every point?

They always add up to the same amount (ME).

Once you know ME at one point, you know it everywhere (because it is conserved within the system).

50,000 J

KE_i + PE_i = KE_f + PE_f

Visual Example of Energy Transformations (90 sec)

Practice

A 65 kg skier starts from rest at the top of a 400 m (vertically) tall hill. Ignore friction.

a. What is the total ME at the top of the hill?

b. What is the total ME at the bottom of the hill?

c. What is the PE and KE at the bottom of the hill?

d. Use the KE equation to find the velocity of the skier at the bottom of the hill.

400 m

top

bottom

Practice

A 65 kg skier starts from rest at the top of a 400 m (vertically) tall hill. Ignore friction.

a. What is the total ME at the top of the hill?

KE_i = 0 ← They start from rest

PE_i = mgh = 65(10)(400) = 260,000 J

ME_i = 0 + 260,000 = 260,000 J

b. What is the total ME at the bottom of the hill?

ME_f = 260,000 J ← It is the same everywhere!

c. What is the PE and KE at the bottom of the hill?

PE_f = 0 ← Height is zero

KE_f = 260,000 J ← All the ME must be KE

d. Use the KE equation to find the velocity of the skier at the bottom of the hill.

KE_f = ½mv_f²

260,000 = 0.5(65)v_f²

v_f = 89.4 m/s

400 m

Practice

A hot dog vendor is pushing their cart at 1.2 m/s on the top of a 15 m high hill when they accidentally let go. The cart rolls all the way down the hill and to the top of the next hill which is only 12 m tall. How fast will the cart be going at the top of the next hill? Ignore friction.

Sketch a bar graph of the ME, KE, and PE of the cart at the top of the first hill and the second hill (relative sizes - no numbers)

first hill

second hill

Practice

A hot dog vendor is pushing their cart at 1.2 m/s on the top of a 15 m high hill when they accidentally let go. The cart rolls all the way down the hill and to the top of the next hill which is only 12 m tall. How fast will the cart be going at the top of the next hill? Ignore friction.

Putting this all in one equation:

ME_i = ME_f

KE_i + PE_i = KE_f + PE_f

½(m)(1.2²) + m(10)(15) = ½(m)(v_f²) + m(10)(12)

0.72m + 150m = 0.5(v_f²)m + 120m

150.72 = 0.5v_f² + 120

v_f² = 61.4

v_f = 7.8 m/s

← All the mass cancels out!

Sketch a bar graph of the ME, KE, and PE of the cart at the top of the first hill and the second hill (relative sizes - no numbers)

External Forces

• ME is conserved if no outside forces do work on the system
• Friction always reduces the amount of ME a system has

ME_i + W_friction = ME_f

What is the ME transformed into by the work done by friction?

Thermal energy (heat), which is a form of non-mechanical energy, (also sound & physical deformation)

Practice

A 10 kg box placed at the top of a 2 m high ramp is released from rest. It slides to the bottom, but in the process friction does -50 J of work on the box. Determine the velocity of the box at the bottom of the ramp and draw the energy bar graphs with values.

top

bottom

Practice

A 10 kg box placed at the top of a 2 m high ramp is released from rest. It slides to the bottom, but in the process friction does -50 J of work on the box. Determine the velocity of the box at the bottom of the ramp and draw the energy bar graphs with values.

ME_i + W_friction = ME_f

KE_i + PE_i + W_friction = KE_f + PE_f

0 + 10(10)(2) - 50 = ½(10)(v_f²) + 0

v_f² = 30

v_f = 5.5 m/s

x

x