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Ch 11 Energy

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Updated 4/28/04

Roller Coaster
WebQuest:   #1 http://www.glencoe.com/sec/science/webquest/content/rollercoast.shtml
  Design        #2  http://www.learner.org/exhibits/parkphysics/coaster/ 
  Design        #3  http://www.angelfire.com/on2/thrillsandchills/

+--------------------------------------------------------------------+
| 746 W = 550 ft*lb/s = 1hp     0.447 m/s = 1 mi/h   1,000 W = 1 kW  |
|  1 mi = 1,610 m               4.45 N    = 1 lb     0.3048 m = 1 ft |
+--------------------------------------------------------------------+

1    a] DEFINE: ENERGY:  The ability to do work.
        POTENTIAL ENERGY:  Stored energy due to an objects postion in a force field.
        KINETIC ENERGY:  Energy due to motion.

     b] DERIVE THE THE EQUATION FOR POTENTIAL ENERGY.
         ΔPE = WORK
             = FxD:  IN A GRAVITATIONAL FIELD F = WEIGHT = mg
             D is in the direction of motion = vertical height = h
        ΔPE = mgΔh
     c] DERIVE THE THE EQUATIONS FOR KINETIC ENERGY
        ΔKE = WORK
              = FxD:  2nd LAW & deffinition of acceleration--->F = m(Vf-Vi)/t
             For uniform acceleration:  D = (Vf+Vi)t/2
        ΔKE = m(Vf-Vi)/t x (Vf+Vi)t/2
             = m(Vf2 - Vi2)/2 = mVf2/2 - mVi2/2
        Thus:  KEf = mVf2/2, KEi = mVi2/2 and in general KE = mV2/2

     d] WHAT ARE THE SYMBOL & UNIT EQUATIONS FOR K.E. & P.E.?
       Symbol:    KE  = mV2/2               PE  = mgh
       Unit:    Joule = kg*m2/s2          Joule = kg*m/s2*m = kg*m2/s2

2      WRITE THE ENERGY CONSERVATION LAW 3 WAYS, EACH IN SYMBOLS & WORDS.
        1]       KE + PA       =  KE' + PE' = CONSTANT
            TOTAL ENERGY BEFORE = TOT ENERGY AFTER
            TOTAL ENERGY IS CONSTANT
        2] SUBTRACT KE + PE FROM BOTH SIDES---> ΔKE + ΔPA  =  0
             MΔV2 + MgΔH =  0
             THE TOTAL CHANGE IN ENERGY = 0
        3] ΔKE = - ΔPE
           THE LOSS OF KINETIC ENERGY = THE GAIN OF POTENTIAL ENERGY
           EX:  A BALL THROWN UPWARD LOOSES K.E. & GAINS P.E.
          OR  THE LOSS OF POTENTIAL ENERGY = THE GAIN OF KINETIC ENERGY
           EX:  A BALL COMES BACK DOWN.
        d] WRITE THE LAW SHOWING HOW ENERGY IS CHANGED IN WORDS AND SYMBOLS
        Work causes a change in potential and kinetic energy.
        W = ΔKE + ΔPE

3      a] DRAW A PENDULUM SHOWING THE ENERGY AT THE HI, MEDIUM, & LOW POINTS.
            Hi:   Energy = PE + KE = mgH + 0        = constant = mgH
        Med:  Energy = PE + KE = mgh + mV2/2    = constant = mgH
        Low:  Energy = PE + KE =  0  + mVmax2/2 = constant = mgH
       b] DERIVE THE EQUATION FOR THE VELOCITY AT THE BOTTOM OF THE SWING.
        mVmax2/2 = mgH, divide by m/2 ---> Vmax2 = 2gH --->Vmax = \/2gH

4      a] STATE THE ASSUMPTION(S) & DERIVE THE EQUATION FOR THE STOPPING DISTANCE 
          RATIO.
          Dh = High Speed Stopping Distance.
          Dl = Low  Speed Stopping Distance.  
          Dh/Dl = ?
          Since the equation with distance is work we will find Wh/Wl
          Wh = ΔKE + ΔPE high   Assume the braking takes place on a flat road.
          Wl   ΔKE + ΔPE low    Thus ΔPE = 0 

          Wh = ΔKE High = KEfinal - KEinital High  
          Wl = ΔKE Low  = KEfinal - KEinital Low 

          Stopping---> KEf = Zero
          Wh = -KEinitial High 
          Wh = -KEinitial Low 
          
          FDh = mVh2/2   Assume the mass of the vehicle is constant.
          FDl   mVl2/2   Assume the force on the break is the same at high speed as 
                         as at low.  The driver does not panic and slam the breaks.

           Dh/Dl = (Vh/Vl)2 

      
       b] FIND THE STOPPING DISTANCE RATIO FOR 60 mph & 30 mph.
           Dh = (Vh/Vl)2 = (60 mi/h)2 = 22 = 4
           Dl              (30 mi/h)2  
                     
       c] GIVE AN EXAMPLE OF THE DISTANCE RATIO FOUND ABOVE.  
           Dh = 4 -----> Dh = 4 Dl,  If Dl = 100 fft, Dh = 400 ft
           Dl

5     A 70 kg TEACHER IS DROPPED OUT A CLASSROOM WINDOW, 4 m HIGH. 
      a] DRAW AND LABEL A DIAGRAM WITH THE EQUATIONS FOR THE TOTAL ENERGY AT EACH 
         METER OF FALL.
                  PE          +   KE  = constant
                 mgh        + mV2/2 = constant

     4 m | 70 kg x 9.8 m/s2 x 4 m + 70 kg (0)2/2 =
         | 2744 Joules            + 0            = 2744 Joules
    -----------------------------------------------------------
     3 m | 70 kg x 9.8 m/s2 x 3 m + ____________ = 2744 Joules
         | 2058 Joules            + ____________ = 2744 Joules
    -----------------------------------------------------------
     2 m | 70 kg x 9.8 m/s2 x 2 m + ____________ = 2744 Joules
         | 1372 Joules            + ____________ = 2744 Joules
    -----------------------------------------------------------
     1 m | 70 kg x 9.8 m/s2 x 1 m + ____________ = 2744 Joules
         |  686 Joules            + ____________ = 2744 Joules
    -----------------------------------------------------------
     0 m | 70 kg x 9.8 m/s2 x 0 m + ____________ = 2744 Joules
         |    0 Joules            + ____________ = 2744 Joules
    -----------------------------------------------------------
    b] FIND HIS SPEED JUST BEFORE CONTACT WITH THE SIDEWALK.
       mV2/2 = 2744 ---> V2 = 2*2744/70 --->t; V = √2*2744/70 = 8.85 m/s

6     A RIFLE WITH A 0.75 m BARREL CAN SHOOT A 4.2 g BULLET AT A SPEED OF 965 m/s. 
      a] DRAW & LABEL THE DIAGRAM WITH THE ASSUMPTION(S) FOR THE BULLET HEIGHT AND FORCE
        After  /|\  V top = 0
                | 
        h = ?   |  Assume no external force, (Ex air friction)
                |
                |   
        Before Vm = 965 m/s. 4.2 g = 0.0042 kg
            After  +---+
                   | G | 
           D=0.75m | u | 
           F = ?   | n | 
           Before  +---+
Note:  Before for Conservation stage = After for Change stage.
    b] DERIVE THE EQUATION FOR THE BULLET HEIGHT
    c] IF SHOT STRAIGHT UP, HOW HIGH (mi) WILL THE BULLET GO?
       Conservation:  KE LOST = PE GAINED
                          mV2 = mgh ------> h =  V2  = (965 m/s)2 = 47,511 m  
                           2        Algebra     2g     2(9.8 m/s2)

                                            h = 47,511 m x   1 mi  = 29.5 mi
                                                           1,610 m
    d] FIND THE AVERAGE FORCE ON THE BULLET, IN lbs?  
       Change:  W = ΔKE + ΔPE
       Since 0.75 m is very small compared to 47,511 m, ΔPE is very small compared to ΔKE
       and can be dropped from the calculation.
          W = ΔKE 
          FD = mV2 
                2 
          F(0.75 m) = 0.0042 kg(965 m/s)2 
                          2
          F(0.75 m) = 1955.6 kg m2/s2 
          F         = 1955.6 kg m2/s2 
                       0.75 m
          F = 2607 kg m/s2  = 2607 N x 1 lb /4.445 N = 586.6 lb

7     A 10 kg TEST ROCKET IS FIRED VERTICALLY.  ITS FUEL CONTAINS 1960 J OF ENERGY.  
      HOW HIGH WILL IT RISE?  

      Total Energy Before = Total Energy After
      Chemical Potential Energy Before = Gravitational Potential Energy After
                                1960 J = mgh
                                h = 1960 J 
                                     mg
                                h =     1960 J        = 20 m
                                    10 kg(9.8 m/s2)

8     A 15 g BULLET (m) IS SHOT INTO A 100 kg WOOD BLOCK (M), HANGING AT THE END OF A  
      STRING, THE WOOD RISES 191 cm. 
      a] DRAW & LABEL THE DIAGRAM.
         Before        After
         h = 0 = PE    V = 0 = KE
         ////////      ////////
             |             \
             |              \
             |               \
             |             +-----+ 
             |             | Wood|
          +-----+          | ►  |
   Bullet | Wood|          +-----+
       ►  |     |             ↨ Δh    
          +-----+ - - - - - - - 

      b] ASSUMPTION(S):  No Air Friction.
         DERIVE THE EQUATION FOR THE BULLET SPEED.
         KE Lost by Bullet = PE Gained by Bullet & Wood                       __________
                       mV2 = (m+M)gΔh ----------> V2 = 2(m+M)gΔh -----> V = √2(m+M)gΔh
                        2         Mult by 2/m              m                       m
      c] FIND THE BULLET'S SPEED-mph  
         m = 15 g x 1 kg/1,000 g = 0.015 kg
         M = 100 kg
         Δh = 191 cm x 1 m/100 cm = 1.91 m
              __________     _____________________
         V = √2(m+M)gΔh =  √2(0.015+100)9.8(1.91) = 500 m/s x   1 mi/h  = 1,117 mi/h
                   m                0.15             0.447 m/s

9    A 15.0 kg MODEL PLANE FLIES HORIZONTALLY AT 12.5 m/s. 
     THE PLANE GOES INTO A DIVE & LEVELS OFF 20.4 m CLOSER TO EARTH.
     a] DRAW & LABEL THE DIAGRAM
           Before                            After  
        +--------+  Vi =12.5 m/s
        |Airplane| ------------>
        +--------+ - - - - - - - - - - - - - 
                                       /|\
                                        |
                                      Δh = 20.4 m
                                        |   +--------+  Vf = ?
                                       \|/  |Airplane| -------->
                             - - - - - - - -+--------+  
     b] DERIVE THE EQUATION FOR THE NEW SPEED OF THE PLANE.
        PE Lost = KE Gained                  
       mgΔh  = m(Vf2 - Vi2) --------------> Vf2 - Vi2 = 2gΔh   
                     2      Multiply by 2/m               __________  
       Vf2 - Vi2 = 2gΔh----->Vf2 = Vi2 + 2gΔh ---->Vf = √Vi2 + 2gΔh

      c] WHAT IS ITS NEW TOTAL VELOCITY. 
              ___________ 
        V = √Vi2 + 2gΔh 
           _______________________  
       V = √12.52 + 2(9.8 m/s2)20.4 m =  23.6 m/s

10    DESIGN A ROLER COASTER WITH A 15 m DIAMETER LOOP (h). 
      a] DRAW & LABEL THE DIAGRAM.
         
         ---------- - - - - - - -
            /|\     \        /|\ 
             |        \       |   Δh
             |         \    _\|/             Vf = ?
             | hi = ?   \  /     \  /|\
             |           \ |Circle|  |   hf = 15 m, R = 7.5 m
            \|/            \_____/  \|/   


      b] ASSUMPTIONS--No external forces/friction.   
      c] FIND THE SPEED NEEDED TO STAY ON THE LOOP. 
         Centripital Force = Weight                __
          mV2 > mg ----------->V2 > gR ----->V >  √gR    
           R       Mult by R/m

      d] FIND THE HEIGHT OF THE FIRST HILL.  
         PE Lost = KE Gained
           mgΔh  = mVf2  ----------->Δh = Vf2 > gR = R  
                    2    Divide by mg     2g    2g   2
         Δh > R/2 = 7.5 m/2 = 3.75 m
         hi = hf + Δh 
         hi > 15 m + 3.75 m 
         hi > 18.75 m 

Extra Problems


1     AN ARCHER PUTS A 0.20 kg ARROW INTO A BOW WITH A 40 lb PULL. 
      AN AVG FORCE OF 20 lb (89 N) IS EXERTED TO DRAW IT BACK 0.6 m. 
      a] DRAW & LABEL THE DIAGRAM.
      b] STATE THE ASSUMPTION(S) & DERIVE THE EQUATION FOR THE ARROW SPEED.
      c] FIND THE SPEED THE ARROW LEAVES THE BOW [mph]. 
      d] STATE THE ASSUMPTION(S) & DERIVE THE EQUATION FOR THE ARROW HEIGHT.
      e] IF THE ARROW IS SHOT STRAIGHT UP, HOW HIGH DOES IT GO?

2    IF WOOD HAD A Hc OF 25 kJ/g AND COST $0.20/g WHILE KEROSENE HAS A Hc OF 35 kJ/g COST  
     $0.40/g, WHICH WOULD BE THE CHEAPER FUEL?
     Cheaper = lower cost/kJ
     Wood:      $0.20/g/25 kJ/g = $0.008/kJ
     Kerosene:  $0.40/g/35 kJ/g = $0.01143/kJ      Wood is cheaper

Notes: 
Rifle bullet speeds vary from 180 to 1,500 m/s (403 to 3,356 mph)
bullets weigh from 200 to 500 grains, (13 to 32 g)
(1 grain = 1/7000 lbs = 0.002286 oz = 0.0648 g, 1 g = 15.43 grains)

Bows come in 35 to 90 lb pulls (156 to 267 N)
Arrows come in 22" to 32" lengths (56 to 81 cm)

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