Answer to Chapter 4

1. The acceleration can be found from the car's one-dimensional motion:

             v = v0 + at;

             0 = [(90km/s)/3.6ks/s)] + a(8.0s), which gives a = -3.13m/s2.

         We apply Newton's second law to find the required average force

              SF = ma;

               F = (1100kg)(-3.13m/s2) =  -3.4 x 103N.

The negative sign indicates that the force is opposite to the velocity.

2. We writeSF = ma from the force diagram for the bucket:

             y-component:  FT - mg = ma;

             63N - (10kg)(9.80m/s2) = (10kg)a;

       Which gives  a = -3.5m/s2 (down)
3. From New ton's their law, the gases will exert a force on the rocket that is equal and opposite to the force the rocket exerts on the gases.

        (a) With up positive, we Write SF = ma from the force diagram for the rocket:

                 Fgases - mg = ma;

                 33 x 106N - (2.75 x 106kg)(9.80m/s2) = (2.75 x 106Kg)a

             which gives a = 2.2m/s2.     

         (b) If we ignore the mass of the gas expelled and any change in g, we can assume a constant acceleration . We find the velocity from

                  v = v0 + at = 0 + (2.2m/s2)(8.0s) =  18m/s.

          (c) We find the time to achieve the height from

                  y = y0 + v0t + 1/2at2;

                 9500m = 0 + 0 + 1/2(2.2m/s2)t2, which gives t = 93s.

       

4 (b) Because the velocity is constant, the acceleration is zero.

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

                      x-component: Fcos q - Ffr = ma = 0, which gives

                      F = (88.0N)cos45° =   62.2N.

         (c)  y-component:  Ffr - mg - Fsin q = ma, which gives

                      Ffr = (14.5kg)(9.80m/s2) + (88.0N)sin45° =   204N.

          (d) We can find the acceleration from the motion of the mower:

                      a = Dv/Dt = (1.5m/s - 0)/(2.5s) = 0.60m/s2.

                For x-component of SF =ma we have

                      Fcos q  - Ffr = ma;

                      Fcos45° - 62.2N = (14.5kg)(0.60m/s2), which give F =  100N.
5. (a) The two crates must have the same acceleration.

              From the force diagram for crate 1 we have

                      x-component: F - F12 - mkFN1 = m1a;

                      y-component: FN1 - m1g = 0, or FN1 =m1g.

               From the force diagram for crate 2 we have

                      x-component: F12 - mkFN2 = m2a;

                      y-component: FN2  - m2g = 0, or FN2 = m2g.

                If we add the two x-equations, we get

                      F- mkm1g - mkm2g = m1a + m2a; 

                      730N - (0.15)(75kg)(9.80m/s2) - (0.15)((110kg)(9.80m/s2) 

                      = (75kg + 110kg)a, which gives  a =  2.5m/s2.

             (b) We can find the force between the crates from the x-equation for crate 2:

                      F12 - mkm2g = m2a;

                      F12 - (0.15)(110kg)(9.80m/s2) = (110kg)(2.5m/s2), which gives

                      F12 =   4.4 x 102N.                         
6. While the roller coaster is sliding down, friction will be up the place, opposing the motion. From the force diagram for the roller coaster, we have SF = ma:

                     x-component: mgsin q - Ffr = mgsin q - mkFN = ma.

                     y-component: FN  - mgcos q = 0.

           We combine these equations to find the acceleration:

                     a = gsin q - mkcos q = g(sin q - mkcos q)

                        = (9.80m/s)[(sin45° - (0.12)cos45°] = 6.10m/s2.

            For the motion of the roller coaster, we find the speed from

                     v2 = v02 +2a(x - x0);

                     v2 = [(60/km/h)(3.6ks/h)]2 + 2(6.10m/s2)(45.0m - 0), which gives  v = 23m/s (85km/h).

 

SEND QUESTIONS BACK TO PROBLEMS