ANSWER TO CHAPTER & HOMEWORK

1.  The minimum work is needed when there is no acceleration.

         (a)  From the force diagram, we write SF = ma:

                        y-component:  FN - mgcosq = 0;

                       x-component:  Fmin - mgsinq = 0.

          For a distance d along the incline, we have

                       Wmin = Fmindcos0° = mgdsinq (1)

                                = (950kg)(9.80m/s2)(310m)sin9.0° =   4.5 x 105J.

           (b)  When there is friction, we have

                        x-component:  Fmin - mgsinq - mkFN = 0,  or

                        Fmin = mgsinq + mkmgcosq,

                   For a distance d along the incline, we have

                        Wmin = Fmindcos0° = mgd(sinq + mkcosq) (1)

                                = (950kg)(9.80m/s2)(310m)(sin9.0° + 0.25cos9.0°) =   1.2 x 106J.

  

 2.      (a)  A × (B + C) = (7.0i - 8.5j) × [(-8.0 + 6.8)i + (8.1 - 7.2)j + 4.2k]

                                     = (7.0)(-12) + (-8.5)(0.9) = -16.1.

          (b)  (A + C) × B = [(7.0 +6.8)i +(-8.5 -7.2)j +0k] × (-8.0i + 8.1j + 4.2k)

                                     =  (13.8)(-8.0) + (-15.7)*(8.1) + 0 =  -238.

          (c)  (B + A) × C = [(-8.0 + 7.0)i + (8.1 - 8.5)j + (4.2 + ))k] × (6.8i - 7.2j)

                                     = (-1.0)(6.8) + (-0.4)(-7.2) + (4.2)(0) =  -3.9.

                                 

3.  The work done in moving the object is the area under the Fx vs. x graph. For the motion from 0.0m to 11.0m, we find the area of two triangles and one rectangle:

                   W = 1/2(300N)(3.0m- 0.0m) +

                              (300N)(7.0m - 3.0M) +

                                   1/2(300N)(11.0m - 7.0m)

                       =   2.3 x 103J.

  

4  For a variable force, we find the work by integration:

                     W = ò F × ds = ò0X (kx + ax3 + bx4)dx

                         = (kx2/2 + ax4/4 + bx5/5) |0X =  kX2/2 + aX4/4 + bX5/5.

5.  On the horizontal FN = mg, so the friction force is Ffr = mmg.

         (a)  The work by the applied force is

                      WF = Fd = (6.0N)(12m) =  72J.

          (b)  The work by friction is

                      Wfr = -FNd = -mmgd = -(0.30)(1.0kg)(9.80m/s2)(12m) =   -35J.

           (c)  The normal force and the weight do no work. The net work increases the kinetic energy of the mass:

                      WF + Wfr = DK;

                      72J - 35J = Kf - 0, which gives Kf =   37J.

6.  (a)  From the force diagram we write SFy = may :

                     FT - mg = ma,

                     FT - (355kg)(9.80m/s2) = (355kg)(0.15)(9.80m/s2),

                which gives FT =  4.00 x 103N.

           (b)  The net work is done by the net force:

                      Wnet = Fneth = (FT - mg)h

                               = [4.00 x 103N - (355kg)(9.80m/s2)](33.0m) =   1.72 x104J.

             (c)  The work done by the cable is

                       Wcable = FTh

                                  = (4.00 x103N)(33.0m) =   1.32 x105J.

             (d0  The work done by gravity is

                        Wgrav = -mgh

                                  = -(355kg)(9.80m/s2)(33.0m) =   -1.15 x 105J.

                      Note that Wnet = Wcable + Wgrav.

              (e)  The net work done on the load increases its kinetic energy:

                         Wnet  = DK = 1/2mv2 - 1/2mvo2;

                          1.72 x 104J = 1/2(355kg)v2 - 0, which gives  v =   9.84m/s.

 

  
7. (a) If we use r as the displacement, the force of gravity is negative and Dr is negative. Thus we can plot the force as positive with a positive change in r, as shown. If we approximate the area as two rectangles,. the average forces for the two are 

at rE + 1/4h = 6.38 x 106m + 1/4(3.0 x 106m) = 7.13 x 106m,

            F1 = GMEm/r2

                 = (6.67 x 10-11N·m2/kg2)(5.98 x 1024kg) (2500kg)/(7.13 x 106m)2

                 = 1.96 x 104N.

at  rE + 3/4h = 6.38 x 106m + 3/4(3.0 x 106m) = 8.63 x 106m,

             F2 = GMEm/rE2 

                  = (6.67 x 10-11N·m2/kg2)(5.98 x 1024kg) (2500kg)/(8.63 x 106m)2

                  = 1.34 x 104N.

From the graph the work done is the area under the F vs. r graph:

             W = (F1 + F2)1/2h

                 = 1/2(1.96 x 104N + 1.34 x 104N)(3.0 x 106m)

                 = 4.95 x 1010J = 5.0 x 1010J.

(b) To find the work by integration, we have

              W = ò Fdr = òre+ hre - (Gmem/r2)dr 

                  = Gmem/r|re+ hre

                  = Gmem[1/re - 1/(re + h)]

                  = Gmem/re[1 - re/(re + h)]

                  = mgre[1 - re/(re + h)]

                  = (2500kg)(9.80m/s2)(6.38 x 106m)(1 - 6.38 x 106m/9.38 x 106m)

                  = 5.00 x 1010J.

Thus we see that our approximation in (a) is with 3%.

 
*8 (a) The work done by gravity is the decrease in the potential energy:

                   Wgrav = -mg(hf - hi)

                             = -(755kg)(9.80m/s2)(0 - 22.5m)

                             = 1.66 x 105J.

    (b) The work done by gravity increases the kinetic energy:

                   Wgrav = DK;

                   1.66 x 105J = 1/2(755kg)v2 - 0, which gives v = 21.0m/s.

   (c) For the motion from the break point to the maximum compression of the spring, we have

                   Wspring + Wgrav = DK; 

                    -(1/2kxf2 - 1/2kxi2) - mg(hf - hi) = 1/2mvf2 - 1/2mvi2;

                    - [1/2(8.00 x 104N/m)x2 - 0] - (755kg)(9.80m/s2)(-x - 22.5m) = 0 - 0.

This is a quadratic equation for x, which has the solution x = -1.95m, 2.13m.

Because x must be positive, the spring compresses 2.13m.

 

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