QUIZ  8 CH. 6 ENERGY/WORK/CONSERVATION OF ENERGY (SAMPLE TEST 3) (ANSWERS) 

VIRTUAL LAB 2  (SEE EXAMPLE 7.7, CHAPTER 7) 

5*, 7*, 8*, 14*, 18*, 19*, 26*, 68*, 33, 78, 82, 85, 86; DON'T FORGET POWER: 48, 56, 57, 49
5*.
(a) Wg = -mgh since the motion is from low to high.   The gravitational work only depends on the vertical distance h.  See DISCUSSION QUESTION 8,  parts a and b. 
(b) No,  but the work done by  the push force lifting the painter does depend on  whether she accelerates up the ladder or not. EXPLAIN
7*. Isolate each block and look ate each force ask  yourself what  work each force does.

Let Block  A be on the table and Block B be hanging vertically.
A:
WN = 0 since the motion is perpendicular to the normal force of magnitude N.
Wg = 0 since the vertical displacement is zero
Wf = friction work = f*D*cos180 = -f*D always for a constant friction force of magnitude f, which you know how to compute.
WT = T*D*cos0  since the tension force of magnitude T is in the same direction as the motion.
B:
WN = 0 since the normal force is zero.
Wg = +mBgh since the motion is from high to low.
Wf = friction work =0 since the friction force is zero.
WT = T*D*cos180  since the tension force of magnitude T is in the opposite direction to the motion.
For each block compute WN  + Wg +  Wf  + WT and you will find it to be zero since the change in kinetic energy is zero.
8*.  Compute  the dot product as shown is section 1.10.
18.*  See the Galilean Lab write up solutions. Had Galileo dropped a watermelon from Pisa, the air resistance force would have been relatively  small since the weight of the object is so large by comparison, but I am not sure it is negligible because of the large surface area. This problem is reasonable since the drop distance of 25.0 m  is relatively short and allows not much time for air resistance effects  to build up as they are often a function of  speed. See section 5.5 on air drag forces.
change in KE = Wg = mgh ( high to low)
Change in KE = (1/2)*m*vf2 - 1/2)*m*vi2  , where " i " means initial and " f " means final.
(c) With air  resistance we have:
 change in KE = Wg = mgh - R*h, where R is the magnitude of the air resistance force assuming it is constant without loss of generality. How would the answers to (a) and (b) change?
19.*  (1/2)*m*vf2 - (1/2)*m*vi2   = total work
(a) total work = +m*g*H is positive since the motion is from high to low when it comes to gravitational work.
(b)  total work = -m*g*H is negative because the motion is from low to high in the case of gravitational  work; note the final speed is zero. 
(c) total work = -f*D, where f is the magnitude of the kinetic friction force and D is the given distance; go back to Ch. 5 and review friction formulae. Final speed is zero. 
(d) Minutia using previous part. Distance D is given. Final speed is not zero, 
(e)  total work = -m*g*H, since the net vertical displacement is upward, when  the gravitational work must be less than zero. Note: You do not need the angle of incline to solve this part.
26.*  The problem is based on  simple proposition: 
(1/2)*m*vf2 - ( 1/2)*m*vi2   = total work = + or - m*g*H, depending on whether the net vertical motion is up or down between the initial and final positions.  APPLY THESE PRINCIPLES TO THIS CASE. 
78.*  Same proposition: 
(i) (1/2)*m*vf2 - (1/2)*m*vi2   = total work = Wg + WF + WN,  work by gravity, the rider's  applied force  and normal force respectively where  last term is zero since  the normal work on the way up slope must be vanish because that force  is perpendicular  to the motion i.e.  makes an angle of 90 degrees. (NOTE: THE SLOPE IS FRICTIONLESS MAKING  WORK OF  FRICTION Wfk  = 0.)

(ii) The total work is the change in kinetic energy: (1/2)*m*vf2 - (1/2)*m*vi2   = total work = WF + Wg  WHERE THE NET VERTICAL DISPLACEMENT IS FROM LOW TO HIGH.  IS GRAVITY WORK POSITIVE OR NEGATIVE? 

(a) Find the change in kinetic energy. FROM THIS INFO  and the proper value of Wg, YOU CAN ANSWER 
PART (b): Find  WF  .     
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