C. Work Homework

Part I Sections 5.1-5.3


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

 Problems:  32 – 42


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:

2.  No, no displacement

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