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Solving Problems with Energy Conservation, continued. Conservation of Mechanical Energy  KE + PE = 0 ( Conservative forces ONLY !! ) or E = KE + PE = Constant For elastic (Spring) PE: PE elastic = (½)kx 2 KE 1 + PE 1 = KE 2 + PE 2

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Conservation of Mechanical Energy

 KE + PE = 0 (Conservative forces ONLY!! )

or E = KE + PE = Constant

  • For elastic (Spring) PE:PEelastic = (½)kx2

KE1 + PE1= KE2+ PE2

 (½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 +(½)k(x2)2

x1 = Initial compressed (or stretched) length

x2 = Final compressed (or stretched) length

v1 = Initial velocity, v2 = Final velocity

slide3

Example 6-11: Toy Dart Gun

Mechanical Energy Conservation

E =

A dart, mass m = 0.1 kg is pressed against the spring of a dart gun. The spring(constant k =250 N/m)is compressed a distance x1 = 0.06 m& released. The dart detaches from the spring it when reaches its natural length (x = 0). Calculate the speed v2it has at that point.

(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)

0 + ( ½ )(250)(0.06)2 = (½)(0.1)(v2)2 + 0

Gives: v2 = 3 m/s

slide4

Example: Pole Vault

Estimate the kinetic energy & the speed required for a 70-kg pole vaulter to just pass over a bar 5.0 m high. Assume the vaulter’s center of mass is initially 0.90 m off the ground & reaches its maximum height at the level of the bar itself.

example 6 12 two kinds of pe

v1 = 0

Example 6-12: Two Kinds of PE

m = 2.6 kg, h = 0.55 m

Y = 0.15 m, k = ?

v2 = ?

A 2 step problem:

Step 1: (a)  (b)

(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2

v1 = 0, y1 = h = 0.55 m, y2 = 0

Gives:v2 = 3.28 m/s

Step 2: (b)  (c) (both gravity & spring PE)

 (½)m(v2)2 + (½)k(y2)2 + mgy2 = (½)m(v3)2 + (½)k(y3)2 + mgy3

y3 = Y = 0.15m, y2 = 0 (½)m(v2)2 = (½)kY2 - mgY

Solve & getk = 1590 N/m

ALTERNATE SOLUTION: (a)  (c) skipping (b)

v3 = 0

example bungee jump
Example: Bungee Jump

m = 75 kg, k = 50 N/m, y2 = 0

v1 = 0, v2 = 0, y1 = h = 15m + y

y =?

Mechanical Energy Conservation

with both gravity & spring (elastic) PE

 (½)m(v1)2 + mgy1

= (½)m(v2)2 + mgy2 + (½)k(y)2

0 + mg(15+y) = 0 + 0 + (½)k(y)2

Quadratic Equation for y:

Solve & get y = 40 m & -11 m

(throw away negative value)

Δy + 15m

(a)  (c) directly!!

other forms of energy energy conservation
Other forms of energy; Energy Conservation

In any process

Total energy is neither created nor destroyed.

  • Energy can be transformed from one form to another & from one body to another, but thetotal amount is constant.

 Law of Conservation of Energy

  • Again:Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only!This is a general Law!!
  • Forms of energy besides mechanical:
    • Heat (conversion of heat to mech. energy & visa-versa)
    • Chemical, electrical, nuclear, ..
slide8
The total energy is neither decreased

nor increased in any process.

  • Energy can be transformed from one form to another & from one body to another, but the

total amount remains constant

 Law of Conservation of Energy

  • Again: Not exactly the same as the Principle of Conservation of Mechanical Energy, which holds for conservative forces only!This is a general Law!!
sect 6 9 problems with friction
Sect. 6-9: Problems with Friction
  • We had, in general:

WNC = KE + PE

WNC = Work done by all non-conservative forces

KE = Change in KE

PE = Change in PE (conservative forces only)

  • Friction is a non-conservative force!So, if friction is present, we have (WNC  Wfr)

Wfr = Work done by friction

Moving through a distance d, friction force Ffrdoes workWfr = - Ffrd

slide10
When friction is present, we have:

Wfr= -Ffrd = KE + PE = KE2 – KE1 + PE2 – PE1

    • Also now, KE + PE  Constant!
    • Instead, KE1 + PE1+ Wfr = KE2+ PE2

or: KE1 + PE1 - Ffrd = KE2+ PE2

  • For gravitational PE:

(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd

  • For elastic or spring PE:

(½)m(v1)2 + (½)k(x1)2 = (½)m(v2)2 + (½)k(x2)2 + Ffrd

slide11

Example 6-13: Roller Coaster with Friction

A roller-coaster car, mass m = 1000 kg, reaches a vertical height of only y = 25 m on the second hill before coming to a momentary stop. It travels a total distance d = 400 m.

Calculate the work done by friction (the thermal energy produced) & calculate the average friction force on the car.

m = 1000 kg, d = 400 m, y1 = 40 m, y2= 25 m, v1= y2 = 0, Ffr= ?

(½)m(v1)2 + mgy1 = (½)m(v2)2 + mgy2 + Ffrd

 Ffr= 370 N

sect 6 10 power
Sect. 6-10: Power

Power Rate at which work is done or rate at which energy is transformed:

  • Average Power:

P = (Work)/(Time) = (Energy)/(time)

  • Instantaneous power:

SI units: Joule/Second = Watt (W) 1 W = 1J/s

British units: Horsepower (hp). 1hp = 746 W

A side note:

“Kilowatt-Hours” (from your power bill). Energy!

1 KWh = (103 Watt)  (3600 s) = 3.6  106 W s = 3.6  106 J

slide13

Example 6-14: Stair Climbing Power

A 60-kg jogger runs up a long flight of stairs in 4.0 s. The vertical height of the stairs is 4.5 m.

a. Estimate the jogger’s power ss output in watts and horsepower.

b. How much energy did this ss require?

slide14
Average Power
  • Its often convenient to write power in terms of force & speed. For example, for a force F & displacement d in the same direction, we know that the work done is:

W = F d So

 F (d/t) = F v = average power

v  Average speed of the object

slide15

Example 6-15: Power needs of a car

Calculate the power required for a 1400-kg car to do the following:a. Climb a 10° hill (steep!) at a steady 80 km/h b. Accelerate on a level road from 90 to 110 km/h in 6.0 s Assume that the average retarding force on the car is FR = 700 N.

a. ∑Fx = 0

F – FR – mgsinθ = 0

F = FR + mgsinθ

P = Fv

l

b. Now, θ = 0

∑Fx = ma

F – FR= 0

v = v0 + at

P = Fv

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