Euler's disk

What energy change occurs below?

 

 

II. Energy (Joules)

 

 

 - 2 kinds

 

 

 

Potential & Kinetic Energy

 

A) Potential Energy - stored energy that an object has due to its position or condition.

 

"At what position(s) or

condition(s) do these

 objects have the

most PE?" 

 

 

 

 

1. Gravitational P.E.g

(PE = Potential Energy)

 

Two factors determine PE 

Fixed Pulley

Source: Elroy M. Avery School Physics (New York: Sheldon and Company, 1895) 141
Copyright: 2009, Florida Center for Instructional Technology. See license.

ETC Clipart 

 

 

 

Height, weight 

 

 

 

P.E.= (mg)h = (weight)h
Joules Kg (m/s2)m N m
    (Not in reference)

 

g on earth = 9.8 m/s2

mg = weight

(force it could exert)

h = height

(distance force could be exerted)

 

 

 

ΔP.E. = mgΔh = wΔh
    (Not in reference)

 

 

Ex 1) What P.E. is gained when a 100. kg object is raised 4.00 m straight up?

Ex 1) What P.E. is gained when a 100. kg object is raised 4.00 m straight up?

m = 100. kg

h = 4.00 m

 

 

ΔP.E. = mgΔh

 

 

= (100. kg)(9.81 m/s2)4.00 m

 

= 3920 J

 

 

 

Ex 2) What PE would be

 gained if an object were

 moved 4 m to the right?

 

 

 

 

ΔP.E. = 0

 

 

 

Ex 3) An object weighing 15. N is lifted from the ground to a height of .22 m.

Find object's change in PE

 

Ex 3) An object weighing 15. N is lifted from the ground to a height of 0.22 m. Find object's change in PE

 

 

ΔP.E. = mgΔh = wΔh
    (Not in reference)

 

w = 15. N

h = .22 m

 

 

ΔP.E. = wΔh

 

ΔP.E. = 15. N(.22m)

 

ΔP.E. = 3.3 Nm

 

 

 

 

"Crow bars reduce the

force needed to move

a body

 

G. P. Quackenbos A.M. A Natural Philosophy: Embracing the Most Recent Discoveries in the Various Branches of Physics, and Exhibiting the Application of Scientific Principles in Every-day Life (New York: D. Appleton and Company, 1859) 95

Clipart ETC

 

 

 

 

"... remember what

you learned on a

seesaw?

Clipart ETC

 

Work = DPE

W + F & D same direction

W - F & D opp. direction

 

Ex) A 1.60 m person lifts a 2.10 kg book from the ground to 2.20 m above the ground.

a) How much work was done lifting the book?

Work = DPE

W = mgDh

Work = 2.10kg[9.80 m/s2]2.20m

 

W = 4.5.3 J

 

 

c) How much W is done relative to ground? / relative to top of head/

 

- AP

 

 

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