|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
(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
|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
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
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