Answers to Chapter 6

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

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

                    y-component: FN - mgcos q = 0;

                   x-component: Fmin -mgsin q = 0.

             For a distance d along the incline, we have

                   Wmin = Fmindcos0° = mgdsin q (1)

                       = (1000kg)(9.80m/s2)(300m)sin17.5°

                       =  8.8 x 105J.

          (b) When there is friction, we have

                    x-component: Fmin - mkgsin q - mkFN = 0, or

                    Fmin = mgsin q + mkmgcos q = 0,

               For a distance d along the incline, we have

                    Wmin = Fmindcos0° = mgd(sin q + mkcos q)(1)

                             = (1000kg)(9.80m/s2)(300m)(sin17.5° + 0.25cos17.5°)

                             =  1.6 x 106J.

 

2. The work done on the arrow increases it kinetic energy:

                W = Fd = DKE = 1/2mv2 - 1/2mv02;

                (95N)(0.80m) = 1/2(0.080kg))v2 - 0, which gives  v = 44m/s.

 

3. (a) With the reference level at the ground, for the potential energy we have

                 DPE = mgDy = (55kg)(9.80m/s2)(3100m -1600m) =  8.1 x 105J.

        (b) The minimum work would be equal to the change in potential energy:

                 Wmin = DPE =  8.1 x 10J.

        (c) Yes, the actual work will be more than this There will be additional work required for any kinetic energy change, and to overcome retarding forces, such as air resistance and ground deformation.

4. (a) For the motion from the bgidge to the lowest point, we use energy conservation:

                 KEi + PEgravi +PEcordi = KEf + PEgravf + PEcordf;

                 0 + 0 + 0 = = + mg(-h) + 1/2k(h - L0)2;

                 0 = -(60kg)(9.80m/s2)(31m) + 1/2k(31m -12m)2,

               which gives k =  1.0 x 102N/m.

          (b) The maximum acceleration will occur at the lowest  point, 

                where the upward restoring force in the cord is maximum:

                   kxmax - mg = mamax;

                   (1.0 x 102N/m)(31m -12m) - (60kg)(9.80m/s2) = (60kg)amax,

              which gives amax =  22m/s2.
5. (a) We find the normal force from the force diagram for the ski:

                   y-component: FN1 = mgcos q;

              which gives the friction force: Ffr1 = mkmgcos q.

              For the work-energy principle, we have

                   WNC = DKE + DPE = (1/2mvf2 - 1/2mvi2) + mg(hf - hi);

                   -mkmgcos q L = (1/2mvf2 - 0) + mg(0 - Lsin q);

                   -(0.090)(9.80m/s2)cos20°(100m) =

                              1/2vf2 - (9.80m/s2)(100m)sin20°,

              which gives vf = 22m/s.

          (b) On the level the normal force is Ffr2 = mg, so the friction force is FN2 = mkmg.

                For the work-energy principle, we have

                     WNC = DKE + DPE = (1/2mvf2 + 1/2mvi2) + mg(hf - hi);

                     - mkmgD = (0 - 1/2mvi2) + mg(0 - 0);

                     - (0.090)(9.80m/s2)D = -1/2(22m/s)2,

                 which gives D =  2.9 x 102m.

6. The work done by the shot-putter increases the kinetic energy of the shot. we find the power from

                       P = W/t = DKE/t = (1/2mvf2 - 1/2mvi2)/t

                          = 1/2(7.3kg)[(14m/s)2 - 0]/(2.0s) =  3.6 x 102W  (about 0.5hp).

 

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