Solving Problems with Energy Conservation, continued
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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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Solving Problems with Energy Conservation, continued

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Solving problems with energy conservation continued

Solving Problems with Energy Conservation, continued


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: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


Solving problems with energy conservation continued

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


Solving problems with energy conservation continued

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, ..


Solving problems with energy conservation continued

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


Solving problems with energy conservation continued

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


  • Solving problems with energy conservation continued

    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


    Solving problems with energy conservation continued

    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?


    Solving problems with energy conservation continued

    • 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


    Solving problems with energy conservation continued

    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/hb. 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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