Answers to Chapter 5

1. If the car does not skid, the friction is static, with Ffr < msFN.

        This friction force provides the centripetal acceleration. we take a coordinate system with the x-axis in the direction of the centripetal acceleration.

        We write SF = ma from the force diagram for the auto:

                      x-component: Ffr = ma = mv2/R.

                      y-component: FN - mg = 0.

        The speed is maximum when Ffr = Ffr,max = msFN.

        When we combine the equations, the mass cancels, and we get

                     msg = vmax2/R;

                     (0.80)(9.80m/s2) = vmax2/(70m), which gives  vmax = 23m/s.

          The mass canceled, so the result is  independent of the mass.

 

2. At the top of the hill, the normal force is upward and the weight id downward, which we select as the positive direction.

        (a) We write SF = ma from the force diagram for the car:

                       mcarg - FNcar = mcarv2/R;

                      (1000kg)(9.80m/s2) - FNcar = (1000kg)(20m/s)2/(100m),

             which gives   F Ncar = 5.8 x 103N.

           (b) When we apply a similar analysis to the driver, we have

                       (70kg)(9.80m/s2) - FNpass = (70kg)(20m/s)2/(100m),

              which gives    FNpass =  4.1 x 102N.

          (c) For the normal force to be equal to zero, we have

                        (1000kg)(9.80m/s2) - 0 = (1000kg)v2/(100m),

               which gives   v= 31m/s  (110km/h or 70mi/h).       

 

3. The acceleration due to gravity on the surface of the neutron star is

                g = F/m = GM/R2 = (6.67 x 10-11N.m2/kg2)(5)(2.0 x1030kg)/(10 x 103m)2 =  6.7 x 1012m/s2.  

 

4. We take the positive direction upward. the spring scale reads the normal force expressed as an effective mass: FN/g.

        We write SF = ma from the force diagram:

                FN - mg = ma, or meffective = FN/g = m(1 + a/g).

        (a) For a constant speed, there is no acceleration, so we have

                meffective = m(1+ a/g) = m =  58kg.

         (b) For a constant speed, there is no acceleration, we have

                meffective = m(1 + a/g) = m =  58kg.

          (c) For the upward (positive) acceleration, we have

                meffective = m(1 + a/g) = m(1 + 0.33g/g) = 1.33(58kg) =  77kg.

          (d) For the downward (negative) acceleration), we have

                meffective = m(1 + a/g) = m(1 - 0.33g/g) = 0.67(58kg) =   39kg.

           (e) In free fall the acceleration is -g, so we have

                 meffective = m(1 + a/g) = m(1 - g/g) =   0.

         

 

SEND QUESTIONS BACK TO PROBLEMS