ANSWER TO CHAPTER 8 HOMEWORK

1.  (a)  Because the force F = (-kx + ax3 + bx4) is a function only of position, it is conservative.

         (b)  We find the form of the potential energy function from

                         U = - ò F × dr = - ò (-kx + ax3 + bx4)i × (dxi + dyj + dzk)

                             = - ò (-kx + ax3 + bx4)dx =  1/2kx2 - 1/4ax4 - 1/5bx5 + constant.<

2.  We choose the potential energy to be zero at the level of the center of mass before the jump (y = )). We find the minimum speed by ignoring any frictional forces. Energy is conserved, so we have

                         E = K1 + U1 = K2 + U2;

                        1/2mv12 + mgy1 = 1/2mv2 + mgy2;

                         1/2mv12 + m(9.80m/s2)(0) = 1/2m(0.70m/s)2 + m(9.80m/s2)(2.10m), which gives v1 =  6.5m/s.

          Note that the initial velocity will not be horizontal, but will have a horizontal component of 0.70m/s.

 3.  We choose y =0 at the lowest point of the swing.

          (a)  We apply conservation of energy from the release point to the lowest point:

                       E = K1 + U1 = K2 + U2;

                       0 + mgL(1- cosqo) = 1/2mva2 + 0;

                       0 + (9.80m/s2)(2.00m)(1-cos30.0°) = 1/2va2,

                which gives va =  2.29m/s.

            (b)   We apply conservation of energy from the release point to the given point:

                       E = K1 + U1 = K3 + U3;

                       0 + mgL(1-cosqo) = 1/2mvb2 + mgL(1 - cosqb);

                       0 + (9.80m/s2)(2.00m)(1 - cos30.0°) 

                          = 1/2 vb2 + (9.80m/s2)(2.00m)(1 - cos15.0°),               

                    which gives vb =   1.98m/s.

             (c)   We apply conservation of energy from the release point to the given point:

                       E = K1 + U1 = K4 + U4;

                       0 + mgL(1 - cosqo) = 1/2mvc2 + mgL(1 -cosqc);

                       0 + (9.80m/s2)(2.00m)(1 -cos30.0°)

                         = 1/2vc2 + (9.80m/s2)(2.00m)[1 -cos(-15.0°)],

                    which gives vc =  1.98m/s.

                    Because this is the same elevation as in part (b), the answer is the same.

              (d) The net force along the cord must provide the radial acceleration:

                       FT - mgcosq  = mv2/L, or  FT = m[(v2/L) + gcosq].   

                    Thus we have

                       FTa = m[(va2/L) + gcosqa] 

                              = (0.070kg){[(2.29m/s)2/(2.00m)] + (9.80m/s2)cos0°} =  0.87N.

                       FTb = m[(vb2/L) + gcosqb] 

                              = (0.070kg){[(1.98m/s)2/(2.00m)] + (9.80m/s2)cos15.0°} =   0.80N.

                       FTc = m[(vc2/L) + gcosqc] 

                               = (0.070kg){[(1.98m/s)2/(2.00m)] + (9.80m/s2)cos15.0°} =   0.80N.

               (e)  With an initial kinetic energy, conservation of energy from the release point to the given point becomes

                        E = K1 + U1 = K4 + U4;

                        1/2mvo2 + mgL(1 -cosqo) = 1/2mv2 + mgL(1 -cosq}.

                      Thus we have

                         1/2(1.20m/s)2 + (9.80m/s2)(2.00m)(1 - cos30.0°) 

                           = 1/2va2 + (9.80m/s2)(2.00m)(1 -cos0°)

                      which gives va =  2.59m/s;

                          1/2(1.20m/s)2 + (9.80m/s2)(2.00m)(1 - cos30.0°) 

                           =  1/2vb2 + (9.80m/s2)(2.00m)(1 - cos15.0°) ,

                      which gives vb =   2.31m/s;

                          1/2(1.20m/s)2 + (9.80m/s2)(2.00m)(1 -cos30.0°)

                            = 1/2vc2  + (9.80m/s2)(2.00m)[1 -cos(-15.0°)].

                      which gives vc =   2.31m/s.

 

4.  On the level the normal force is FN = mg, so the friction force is Ffr = mkmg.

         For the work-energy principle, we have

                         WNC = DK + DU = (1/2mvf2 - 1/2mvi2) + mg(hf - hi);

                         F(L1 + L2) - mkmgL2 = (1/2mvf2 - 0) + mg(0 - 0);

                         (350N)(15m + 15m) - (0.25)(90kg)(9.80m/s2)(15m) = 1/2(90kg)vf2,

         which gives vf =   13m/s.

 

5.  The escape velocity from a mass M, as determined by energy conservation, is from

                          vesc2 = 2GM/r.

         (a)  To escape from the the Sun's surface, we have

                          vesc2 = 2GMS/rS

                                   = 2(6.67 x 10-11N.m2/kg2)(2.0 x 1030kg)/(7.0 x 108m),

                 which gives vesc =  6.2 x 105m/s.

          (b)  To escape from the Sun when at the Earth's location, we have

                           vesc2 = 2GMS/r

                                    = 2(6.67 x 10-11N.m2/kg2)(2.0 x 1030kg)/(1.50 x 1011m),

                 which gives vesc =   4.2 x 104m/s.

                 Because the gravitational attraction provides the radial acceleration of the Earth, we have

                           GMSME/r2 = MEvorbit2/r, or v orbit2 = GMS/r.

                      for the ratio we get  = 

                            vesc2/vorbit2 = (2GMS/r)/(GMS/r), or vesc/vorbit = Ö2.

6.  We find the average resistance force from the acceleration:

                            FR = ma = mDv/Dt = (1000kg)(70km/h - 90km/h)/(3.6ks/h)(60s) = -926N.

          If we assume that this is the resistance force at 80km/h, the engine must provide an equal and opposite force to 

          maintain a constant speed. We find the power required from

                            P = Fv = (926N)(80km/h)/(3.6ks/h) =  2.1 x 104W = (2.1) x 104W)/(746W/hp) =  28hp.     

*7 (a) The net force toward the center of the sphere provides the radial acceleration:

                           mgcosq - FN = mv2/r.

The skier will leave the sphere when the normal force becomes zero, or

                           va2 = grcosqa.

With the reference level at the center of the sphere, we apply conservation of energy from the top to the point where the skier leaves:

                           E = K1 + U1  = K2 + U2;

                           0 + mgr = 1/2mva2 + mgrcosqa;

                           gr = 1/2grcosqa + grcosqa, which gives cosqa = 2/3, qa = 48.2°

(b) the condition for the normal force to become zero is 

                           vb2 = grcosqb.

There will be negative work done by the friction force, so we have

                           Wf = DK + DU  = (1/2mvb2 - 0) + mgr(cosqb - 1)

                                = 1/2mgrcosqb + mgrcosqb - mgr, 

which we write

                           cosqb = 2/3[1 + (Wf/mgr)].

Because Wf  is negative, this means that

                           cosqb < cosqa, or  qb > qa.

 
*8. Because the rate of work is P = Fv and the applied force produces the acceleration, we find the velocity and acceleration as a function of time:

                            x = (5.0m/s3)t3 - (8.0m/s2)t2 - (30m/s)t;

                            v = dx/dt = (15.0m/s3)t2 - (16.0m/s2)t - (30m/s);

                            a = dv/dt = (30.0m/s3)t - (16.0m/s3).

Thus the rate of work is

                           P = Fv = mav = m[(30.0m/s3)t - (16.0m/s2)][(15.0m/s3)t2 - (16.0m/s2)t - ((30m/s)].

(a) At t = 2.0s, we have

                           

                           P = (0.280kg)[(30.0m/s3)(2.0s) - (16.0m/s2)][(15.0m/s3)(2.0s)2 - (16.0m/s2)(2.0s) - (30m/s))]

                               = - 25W.

(b) At t = 4.0s, we have

                           P = (0.280kg)[(30.0m/s3)(4.0s) - (16.0m/s2)][(15.0m/s3)(4.0s)2 - (16.0m/s2)(4.0s) - (30m/s))]

                               = + 4.3 x 103W.

(c) Over a time interval, the average net power produces the change in kinetic energy:

                           P = W/t = DK/Dt = (1/2mvf2 - 1/2mvi2)/Dt = 1/2m(vf2 - vi2)/Dt.

We find the velocities at the three times:

                            v0 = (15.0m/s3)(0)2 - (16.0m/s2)(0) - (30m/s) = - 30m/s;

                            v2 = (15.0m/s3)(2.0s)2 - (16.0m/s2)(2.0s) - (30m/s) = - 2.0m/s;

                            v4 = (15.0m/s3)(4.0s)2 - (16.0m/s2)(4.0s) - (30m/s) = 146m/s.

From t = 0 to t = 2.0s, we have

                            P = 1/2(0.280kg)[(-2.0m/s)2 - (-30m/s)2]/(2.0s - 0) = -63W.

From t = 2.0s to t = 4.0s, we have

                            P = 1/2(0.280kg)[(146m/s)2 - (-2.0m/s)2]/(4.0s - 2.0s) = + 1.5 x 103W.

 

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