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Conservation of Energy by mikesanye

VIEWS: 8 PAGES: 5

									CHAPTER 8: Conservation of Energy

Solutions to Problems

22. (a)    Draw a free-body diagram for each block. Write
                                                                                                          FN                     FT
          Newton’s second law for each block. Notice that the                                                   FT          yB
          acceleration of block A in the yA is 0 zero.                                      yA
                                                                                                 xA                              mB g
                Fy1  FN  mA g cos  0  FN  mA g cos                                            
                                                                                                           

                                                                                                           mAg

           F   x1
                      FT  mA g sin   mAaxA


               F     y2
                            mB g  FT  mBa yB  FT  mB  g  a yB  Since the blocks are


          connected by the cord, a yB  a xA  a.                      Substitute the expression for the tension

          force from the last equation into the x direction equation for block 1, and solve for
          the acceleration.

               mB  g  a   mA g sin   mA a  mB g  mA g sin   mA a  mBa
                          m                    
                                                       9.80 m s  
                                    mA sin                             5.0 kg  4.0 kg sin 32 
               ag                                                                                     3.1m s 2
                               B                                   2

                             mA  m B                                          9.0 kg

    (b) Find the final speed of m B (which is also the final speed of mA ) using constant
          acceleration relationships.




    v 2  v0  2ay  v 2  2 g
           2                                 m  B
                                                       mA sin    h     
      f                 f
                                                  mA  mB 
                 m                     
                                                                     0.75 m  
                           mA sin                                                 5.0 kg  4.0 kg sin 32 
    v f  2 gh
                      B

                      mA  mB 
                                                  
                                                2 9.80 m s 2
                                                                                            9.0 kg
                                                                                                                 2.2 m s



    (c) Since there are no dissipative forces in the problem, the mechanical energy of the
          system is conserved. Subscript 1 represents the blocks at the release point, and
          subscript 2 represents the blocks when m B reaches the floor. The ground is the
          zero location for gravitational potential energy for mB , and the starting location for
          mA is its zero location for gravitational potential energy.                        Since m B falls a
          distance h, mA moves a distance h along the plane, and so rises a distance h sin .
          The starting speed is 0.
                 E1  E2  0  mA gh               1
                                                    2    mA  mB  v  2
                                                                       2
                                                                            mB gh sin  

                                mB  mA sin  
                  v2  2 gh                   
                                mA  mB 
      This is the same expression found in part (b), and so gives the same numeric result.


36. (a)      Use conservation of energy to equate the potential energy at the top of the circular
     track to the kinetic energy at the bottom of the circular track. Take the bottom of the
     track to the be 0 level for gravitational potential energy.

                   Etop  Ebottom  mgr  1 mvbottom 
                                          2
                                              2




                                             
                   vbottom  2 gr  2 9.80 m s2                   2.0 m   6.261m s  6.3m s
      (b)    The thermal energy produced is the opposite of the work done by the friction force.
In this situation, the force of friction is the weight of the object times the coefficient of kinetic
friction.


      E thermal  Wfriction   Ffriction x   Ffriction x cos    k mg x  cos180   k mg x

                                     
               0.251.0 kg  9.80 m s 2           3.0m   7.35J  7.4 J
     (c) The work done by friction is the change in kinetic energy of the block as it moves
            from point B to point C.

                   Wfriction  K  KC  K B  1 m vC  vB
                                               2
                                                    2    2
                                                                          

                           2Wfriction               2  7.35J 
                   vC                    vB                        6.261m s   4.9498 m s  4.9 m s
                                            2                                      2

                               m                        1.0 kg 
     (d) Use conservation of energy to equate the kinetic energy when the block just
            contacts the spring with the potential energy when the spring is fully compressed
            and the block has no speed. There is no friction on the block while compressing
            the spring.

                   Einitial  Efinal       1
                                            2
                                                mvcontact  1 kxmax 
                                                  2
                                                            2
                                                                2


                           2
                          vcontact                  4.9498 m s  2
                   k m               1.0 kg                         612.5 N m  610 N m
                            2
                           xmax                        0.20 m  2
74. First, consider a free-body diagram for the cyclist going down hill. Write
                                                                                                 Ffr        FN
    Newton’s second law for the x direction, with an acceleration of 0 since the                                       y
    cyclist has a constant speed.
                                                                                                                           x
                                                                                                       
                                                                                                       
         F     x
                     mg sin   Ffr  0  Ffr  mg sin 
                                                                                                            mg
                                                                                                                   




                                                                                                  y
    Now consider the diagram for the cyclist going up the hill. Again, write
                                                                                                       x
    Newton’s second law for the x direction, with an acceleration of 0.                                      FN
                                                                                                 FP

         F
                                                                                                                 Ffr
                x
                     Ffr  FP  mg sin   0  FP  Ffr  mg sin 
                                                                                                       
    Assume that the friction force is the same when the speed is the same, so the                           mg
                                                                                                                   
    friction force when going uphill is the same magnitude as when going
    downhill.
          FP  Ffr  mg sin   2mg sin 
    The power output due to this force is given by Eq. 8-21, with the force and velocity
    parallel.

                                               
          P  FPv  2mgv sin   2  75 kg  9.80 m s2        4.0 m s sin 6.0   o
                                                                                        610 W
90. (a)      Draw a free-body diagram for the block at the top of the curve.             Since
the
                                                                                                         FN    mg
            block is moving in a circle, the net force is centripetal. Write Newton’s
            second law for the block, with down as positive.             If the block is to be on
            the verge of falling off the track, then FN  0.

                 F      R
                              FN  mg  mv 2 r  mg  mvtop r  vtop  gr
                                                         2




            Now use conservation of energy for the block. Since the track is frictionless, there
            are no non-conservative forces, and mechanical energy will be conserved.
            Subscript 1 represents the block at the release point, and subscript 2 represents the
            block at the top of the loop. The ground is the zero location for potential energy

             y  0 .       We have v1  0, y1  h, v2  gr , and y2  2 r.            Solve for h.


                  E1  E2               1
                                         2
                                             mv12  mgy1  1 mv2  mgy2  0  mgh  1 mgr  2mgr 
                                                           2
                                                               2
                                                                                    2

                 h  2.5 r

      (b)    See the free-body diagram for the block at the bottom of the loop. The net
            force is again centripetal, and must be upwards.                                                  FN

                 F      R
                              FN  mg  mv 2 r  FN  mg  mvbottom r
                                                              2
                                                                                                                    mg

            The speed at the bottom of the loop can be found from energy conservation, similar
            to what was done in part (a) above, by equating the energy at the release point
            (subscript 1) and the bottom of the loop (subscript 2). We now have v1  0,
            y1  2h  5r , and y2  0.                 Solve for v 2 .

             E1  E2              1
                                   2
                                       mv12  mgy1  1 mv2  mgy 2  0  5mgr  1 mvbottom  0 
                                                     2
                                                         2
                                                                                2
                                                                                    2



             vbottom  10 gr  FN  mg  m vbottom r  mg  10mg  11mg
              2                             2




      (c)    Again we use the free body diagram for the top of the loop, but now the normal
force does not
            vanish. We again use energy conservation, with v1  0, y1  3r , and y2  0.
      Solve for v 2 .

                   F        R
                                  FN  mg  m v 2 r  FN  m vtop r  mg
                                                               2



                   E1  E2                  1
                                             2
                                                 mv12  mgy1  1 mv2  mgy2  0  3mgr  1 mvtop  0 
                                                               2
                                                                   2
                                                                                         2
                                                                                             2




                   vtop  6gr  FN  mvtop r  mg  6mg  mg  5mg
                    2                  2




      (d)    On the flat section, there is no centripetal force, and FN  mg .

								
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