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Chapter 6, Continued. Summary so Far. Work (constant force): W = F || d =Fd cos θ. Work-Energy Principle: W net = (½)m(v 2 ) 2 - (½)m(v 1 ) 2  KE Total work done by ALL forces!. Kinetic Energy: KE  (½)mv 2. Sect. 6-4: Potential Energy.

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summary so far
Summary so Far
  • Work (constant force):
  • W = F||d =Fd cosθ
  • Work-Energy Principle:
  • Wnet = (½)m(v2)2 - (½)m(v1)2 KE
  • Total work done by ALLforces!
  • Kinetic Energy:
  • KE  (½)mv2
slide3

Sect. 6-4: Potential Energy

A mass can have aPotential Energydue to its environment

Potential Energy (PE) 

Energy associated with the position or configuration of a mass.

Examples of potential energy:

A wound-up spring

A stretched elastic band

An object at some height above the ground

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Potential Energy (PE)  Energy associated with the position or configuration of a mass.

Potential work done!

Gravitational Potential Energy:

PEgrav mgy

y = distance above Earth

m has the potential to do work

mgy when it falls

(W = Fy, F = mg)

slide5

Gravitational Potential Energy

We know that for constant speed

ΣFy = Fext – mg = 0

So, in raising a mass m to a height h, the work done by the external force is

Fexthcosθ

So we define the gravitational potential energy at a height y above some reference point (y1) as

(PE)grav

slide6
Consider a problem in which the height of a mass above the Earth changes from y1to y2:
  • The Change in Gravitational PE is:

(PE)grav= mg(y2 - y1)

  • Work done on the mass: W = (PE)grav

y = distance above Earth

Where we choose y = 0 is arbitrary, since we take the difference in 2 y’s in (PE)grav

slide7

Of course, thispotential energy can be converted to kinetic energy if the object is dropped.

Potential energy is a property of a system as a whole, not just of the object (because it depends on external forces).

If PEgrav = mgy, from where do we measure y?

It turns out not to matter,

as long as we are consistent about where we choose y = 0. Because only changes in potential energy can be measured.

slide8

Example 6-7: Potential energy changes for a roller coaster

A roller-coaster car, mass m = 1000 kg, moves from point 1 to point 2 & then to point 3.

∆PEdepends only on

differences in vertical

height.

a. Calculate the gravitational potential energy at points 2 & 3 relative to point 1. (That is, take y = 0 at point 1.)b. Calculate thechangein potential energy when the car goes from point 2 to point 3. c. Repeat parts a. & b., but take the reference point (y = 0) at point 3.

slide9

Many other types of potential energy besides gravitational exist!

Consider an IdealSpring

AnIdeal Spring, is characterized by

a spring constant k, which is a measure

of it’s “stiffness”. The restoring force

of the spring on the hand:

Fs = - kx

(Fs >0, x <0; Fs <0, x >0)

This is known as Hooke’s “Law”(but, it isn’t

really a law!)It can be shown that the

work done by the person is

W = (½)kx2 (PE)elastic

We use this as the definition of

Elastic Potential Energy

slide11

Elastic Potential Energy (PE)elastic≡(½)kx2

Relaxed Spring

The work to compress the spring a distance x is

W = (½)kx2 (PE)elastic

The spring stores potential energy!

When the spring is released,

it transfers it’s potential energy

PEe = (½)kx2to the mass in the

form of kinetic energy

KE = (½)mv2

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In a problem in which compression or stretching distance of spring changes from x1 to x2.
  • The change in PE is:

(PE)elastic= (½)k(x2)2 - (½)k(x1)2

  • The work done is: W = - (PE)elastic

The PE belongs to the system,

not to individual objects

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