C. Work
Homework
5.1 Work done by a constant
force
Problems: 1-11, 13-15
Helpful
Tips:
5: Work changes during flight. Draw a circle and remember that
W =
(Fcosq)d.
So, as the angle changes, so does Work.
6: Force in the x direction where q = 0 will be lowest
7: The answer will be negative because friction
acts opposite to motion
8. W = mgh
10. To find F,
use mgsinq, NOT mgcosq. Draw it out and you’ll see why
11. subtract Fx - Fmk to find Fnet. W = Fnetd
13: w = mgh
14: Megan Loves
this problem! Dad is REDUCING the
normal force since he is pulling UP on the sled at Fsinq.
Therefore, he is reducing the Normal force by Fsinq.
Realize that FCosq is the only component of Dad’s motion that
is doing any pulling (because the sled only moves in the x direction, It doesn’t move into the mountain or away
from it in the y direction). When FCosq is equal to or greater than Fk,
then she moves! Since the problem says
she is moving with a constant velocity,
The force that dad pulls in the x direction (FCosq) must be equal to but
opposite from the force of friction of the hill (Fk). Therefore, the sled is moving but with no
acceleration. Mathematically, we
represent this as: Fk -
FCosq = ma where a = 0 m/s2. So, solve for Fk and you have
solved for FCosq.
Fk = mk-N. And since N = mg - Fsinq, simply substitute mg - Fsinq for N to
solve for Fk.
15: Megan
Loves this problem too! She
says: “Draw a free-body diagram so that
it makes more sense to you”. Dad is
pushing Horizontally with respect to the hill.
Remember the hill is in the X direction. Therefore, Dad’s force is 15o BELOW the x axis. See
that? That means that Dad’s force (Fdad)
is ADDING to the Normal force by Fdadsinq. N =
mgCosq + FdadSinq
(mgCosq is Megan’s y-component mass
falling down into the hill along the y-axis only).
Realize that Megan moves up the hill
because of Dad’s force in the x direction (FdadCosq) MINUS
her body force in the x direction (mmeganandsledgSinq) MINUS
the force of kinetic friction in the x direction
(Fk
= mkN). If you draw the free body diagram (as Megan
suggested to you earlier), you will clearly see that Dad is pushing up the hill
while Megan’s mass and Fk are both working against him. That’s why you subtract those forces from
his FdadCosq.
åFxdirection = FdadCosq - mgsinq - Fk
(remember that Fk = mkN and remember that N =
mgCosq + FdadSinq.
Perform all necessary substitutions before plugging in any numbers)
5.2 Springs
Problems: 19, 20, 22, 23, 24, 26, 28, 29, 31
Helpful
Tips:
20: Work = 1/2 K (x2 – xo2). Assume that K = 1 to make it easy – it
doesn’t matter what value you use for K.
For W1, xo = 0 and x = 2 cm. For W2, Xo = 2 cm and
x = 6 cm.
22: Convert cm to m
24. W = 1/2KDx2
31: remember to subtract the areas.
5.3 Work Energy Theorem: (refer to these equalities as you do the problems)
Worknet = 1/2 mv2 –
1/2 mvo2 = DKinetic Energy = (FCosq)d
32: W = FCosq d. And, if q is greater than 90o,
then Cos q is negative – less than
zero.
35: KE = 1/2 mv2. Find KE of the original condition and then
subtract that from KE of the “bounce”.
36: Refer to the work-energy theorem equalities
above AND realize that
Wnet = Fk d = mkNd = mkmgd.
Thererfore, 1/2mv2
= mkmgd. Notice that MASS CANCELS!!! Yippeeee!
39: Wnet = DKE = FCosqd
40: Convert g to Kg and cm to m
41: Wnet = 1/2mv2 – 1/2mvo2
42: Realize V = 0 m/s because the car
stops. Therefore, Wnet = 0 –
1/2mvo2
Even Answers:
4. zero.
q = 90 degrees. Cos of 90 is zero
6. 5.0 N
8. 1.5 x 103 J
10. 15.1 J, gravitational force
14. 6.1 x 102J
20. d
22. 80 N/m
24. 8.0 x 10-3J
26. 0.64 m
28. a)
4.5 J b) 3.5 J
32. b
34. a
36. The same
38. a)
3.8 x 105 J b) -3.8 x 105 J
40. 1.5 x 103 N
42. 200 m