HK 7
pg.176-5,6,25*,33*, 40,48,73,77*

5
a)  0  weight is a conservative force - work done on closed loop (initial and final points are at the same height) must be zero
b)  mg*h/2
c)  mgh
 

 a)  mgL      It doesn't matter what is the path: a or b.
6.
 b)  -mgL      It doesn't matter what is the path: a or b. 
c)  0   the path is closed loop from weight point of view.  Started and finished at the same level.
d)   -mgL,   +mgL,   0

40.
  a)   +T-688=69v2/18 (Newton's Law)    and
work done by weight  688*3.2 = 1/269*v2 Kinetic Energy.
Solve for v, then calculate T.
b)  if it breaks at an angle (means that the tension is 950 N) :
950cosQ -688 = 69v2/18
work done by weight  688*(3.2-18+18cosQ) = 1/269*v2  Kinetic Energy. 
and solve for v and Q .

48.
F(x) = -dU(x)/dx                                                                                                                                                         graph of U(x)

a)   -dU(x)/dx = -[-17-(-2)]/(4-1) = 5 N.  Force is positive so it points in positive direction of x axis.
c)  at x = 2    v = -1.5 m/s   and U = -7 (see graph).  At x = 7 U = -17(see graph).  DU = -17-(-7)=-10
and DU+DK=0     so    DK=Kfinal-Kinitial=-DU = 10    and Kinitial=1/22*(1.5)2 so calculate Kfinal then vfinal2 then vfinal .

73.
a)    Potential Energy = 1/2320*(0.075)2 = - work done by the spring
b)    work done by friction force = -Nmk*0.075  (friction force * distance*cos180)    Normal force N = 25 Newtons.
c)    Initial Energy = Kinetic Energy = 1/22.5*(v)2
       Final Energy = Potential Energy = 1/2320*(0.075)2
       |work| done by friction force = Nmk*0.075  (friction force * distance)     Normal force N = 25 Newtons.
    1/22.5*(v)2 = 1/2320*(0.075)2+Nmk*0.075   and solve for v