**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 F_{x} - F_{m}_{k} to find F_{net}. W = F_{net}d

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 F_{k},
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 (F_{k}). Therefore, the sled is moving but with __no
acceleration__. Mathematically, we
represent this as: F_{k} -
FCosq = ma where a = 0 m/s^{2}. So, solve for F_{k} and you have
solved for FCosq.

F_{k} = m_{k}-N. And since N = mg - Fsinq, simply substitute mg - Fsinq for N to
solve for F_{k}.

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 15^{o} **BELOW** the x axis. See
that? That means that Dad’s force (F_{dad})
is ADDING to the Normal force by F_{dad}sinq. N =
mgCosq + F_{dad}Sinq
(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 (F_{dad}Cosq) **MINUS**
her body force in the x direction (m_{meganandsled}gSinq) **MINUS**
the force of kinetic friction in the x direction

(F_{k}
= m_{k}N). 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 F_{k} are both working against him. That’s why you subtract those forces from
his F_{dad}Cosq.

åF_{xdirection} = F_{dad}Cosq - mgsinq - F_{k}

(remember that F_{k} = m_{k}N and remember that N =
mgCosq + F_{dad}Sinq.
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 (x^{2} – x_{o}^{2}). Assume that K = 1 to make it easy – it
doesn’t matter what value you use for K.
For W_{1}, x_{o} = 0 and x = 2 cm. For W_{2}, X_{o} = 2 cm and
x = 6 cm.

22: Convert cm to m

24. W = 1/2KDx^{2}

31: remember to subtract the areas.

5.3 Work Energy Theorem: (refer to these equalities as you do the problems)

Work_{net} = 1/2 mv^{2} –
1/2 mv_{o}^{2} = DKinetic Energy = (FCosq)d

32: W = FCosq d. And, if q is greater than 90^{o},
then Cos q is negative – less than
zero.

35: KE = 1/2 mv^{2}. 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

W_{net} = F_{k} d = m_{k}Nd = m_{k}mgd.

Thererfore, 1/2mv^{2}
= m_{k}mgd. Notice that MASS CANCELS!!! Yippeeee!

39: W_{net} = DKE = FCosqd

40: Convert g to Kg and cm to m

41: W_{net} = 1/2mv^{2} – 1/2mv_{o}^{2}

42: Realize V = 0 m/s because the car
stops. Therefore, W_{net} = 0 –
1/2mv_{o}^{2}

**Even Answers:**

4. zero.
q = 90 degrees. Cos of 90 is zero

6. 5.0 N

8. 1.5 x 10^{3} J

10. 15.1 J, gravitational force

14. 6.1 x 10^{2}J

20. d

22. 80 N/m

24. 8.0 x 10^{-3}J

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 10^{5} J b) -3.8 x 10^{5} J

40. 1.5 x 10^{3} N

42. 200 m