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UNIT 4 Work, Energy, and PowerPowerPoint Presentation

UNIT 4 Work, Energy, and Power

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### Thursday November 10th

UNIT 4Work, Energy, and Power

1) 2 v1 = v2

2) 2 v1 = v2

3) 4 v1 = v2

4) v1 = v2

5) 8 v1 = v2

ConcepTest 6.5bKinetic Energy IICar #1 has twice the mass of car #2, but they both have the same kinetic energy. How do their speeds compare?

1) 2 v1 = v2

2) 2 v1 = v2

3) 4 v1 = v2

4) v1 = v2

5) 8 v1 = v2

ConcepTest 6.5bKinetic Energy IICar #1 has twice the mass of car #2, but they both have the same kinetic energy. How do their speeds compare?

Since the kinetic energy is 1/2 mv2, and the mass of car #1 is greater, then car #2 must be moving faster. If the ratio of m1/m2 is 2, then the ratio of v2 values must also be 2. This means that the ratio of v2/v1 must be the square root of 2.

CONSERVATION OF MECHANICAL ENERGY

Thursday, November 10

- Conservation of Mechanical Energy
- Hw: Practice D (All) p172

UPCOMING…

- Fri: More Conservation of Energy:
- Bowling Ball Demo
- Power
- Mon: Problem Quiz 1
- Tue: Problems @ the Boards

Section 3 Conservation ofEnergy

Chapter 5

Mechanical Energy- Mechanical energyis the sum of kinetic energy and all forms of potential energy associated with an object or group of objects.
ME = KE + ∑PE

- Mechanical energy is often conserved.
MEi = MEf

initial mechanical energy = final mechanical energy (in the absence of friction)

Section 3 Conservation ofEnergy

Chapter 5

Sample ProblemConservation of Mechanical Energy

Starting from rest, a child zooms down a frictionless slide from an initial height of 3.00 m. What is her speed at the bottom of the slide? Assume she has a mass of 25.0 kg.

Section 3 Conservation ofEnergy

Chapter 5

Sample Problem, continuedConservation of Mechanical Energy

1. Define

Given:

h = hi = 3.00 m

m = 25.0 kg

vi = 0.0 m/s

hf = 0 m

Unknown:

vf = ?

Section 3 Conservation ofEnergy

Chapter 5

Sample Problem, continuedConservation of Mechanical Energy

3. Calculate

Substitute values into the equations:

PEg,i = (25.0 kg)(9.81 m/s2)(3.00 m) = 736 J

KEf = (1/2)(25.0 kg)vf2

Now use the calculated quantities to evaluate the final velocity.

MEi = MEf

PEi + KEi = PEf + KEf

736 J + 0 J = 0 J + (0.500)(25.0 kg)vf2

vf = 7.67 m/s

Section 3 Conservation ofEnergy

Chapter 5

Mechanical Energy, continued- Mechanical Energy is not conserved in the presence of friction.
- As a sanding block slides on a piece of wood, energy (in the form of heat) is dissipated into the block and surface.

Systems and Energy Conservation

Ball dropped from rest falls freely from a height h.

Find its final speed.

h

v

Energy

Systems and Energy Conservation

A block of mass m compresses a spring (force constant k) a distance x. When the block is released, find its final speed.

v

m

m

x

Energy

Systems and Energy Conservation

m

When released from rest, the block slides to a stop.

Find the distance the block slides.

vf= 0

Friction (m)

m

k

x

d

Energy

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