ANSWER TO CHAPTER 4 HOMEWORK

1.  The requited average acceleration can be found from the one-dimensional motion:

                 v2 = vo2 + 2a(x - xo);

                0 = [(100km/h)(3.6ks/h)]2 + 2a(150m - 0), which gives a = -2.57m/s2.

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

                SF = ma;

                 F = (3.6 x105kg)(-2.57m/s2) = -9.3 x 105N.

         The weight of the train is

                mg = (3.6 x105kg)(9.80m/s2) = 3.5 x 106N,

         so Superman must apply a force that is 25% of the weight of the train.

2.  we write SF = ma from the force diagram for the car

               y-component : F^ - mg = ma, or

              F^ = m(a + g) = (1200kg)(0.80m/s2 + 9.80m/s2)

                    = 1.3 x 104N.

 
3.  From Newton's third 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 x106N - (2.75 x106kg) (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 = vo + at = 0 + (2.2m/s2)(8.0s) = 18m/s.

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

                 y = yo + vot + 1/2at2;

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

                  

 

  
4.  (a)  We select the helicopter and the car as the system.

                We write SF = ma from the force diagram:

                    Fair - mhg - mcg = (mh + mc)a, which gives

                    Fair = (mh + mc)(a + g)

                           = (7500kg + 1200kg)(0.52m/s2 + 9.80m/s2)

                           = 8.98 x 104N.

           (b) we select the car as the system.

                We write SF = ma from the force diagram:    

                    FT - mcg = mca, which gives 

                   FT = mc(a + g)

                         = (1200kg)(0.52m/s2 + 9.80m/s2)

                        = 1.24 x 104N.

                 

 		 
5.   There is no acceleration. From the symmetry of the force diagram, we see that the tension is the same on each side of Arlene.

           We write SF = ma from the force diagram: 

                   SFy = may:  2FTsinq - mg = 0, which gives  

                     FTmin = mg /2sinqmax 

                                = (50.0kg)(9.80m/s2)/2sin10° = 1.4 x103N.

  
6. (a) From the force diagram for the block, we have SF = ma:

                     x-component:   mgsinq  = ma;

                     y-component:   FN - mgcosq = 0

              From the x-equation we find the acceleration:

                     a = gsinq = (9.80m/s2)sin22.0° = 3.67m/s2.

             (b) For the motion of the block, we find the speed from 

                     v2 = vo2 + 2a(x - xo );

                     v2 = 0 + 2(3.67m/s2)(12.0m - 0), which gives v = 9.39m/s.

 

 
7.  (a) See diagram.

         (b) If the system is released from rest, the blocks will have the same acceleration in the directions indicated on the diagram. We write SF = ma from the force diagram for each block:         

                    x-component (m1):  FT = m1a;

                    y-component (m1):  FN  - m1g = 0;

                    y-component (m2):  m2g - FT = m2a.

             When we add the first and third equations, we get

                    m2g = (m1 + m2)a, which gives a = m2g/(m1 + m2).

              When we use this result in the first equation, we get

                    FT = m1m2g/(m1 + m2).

  
8*. Forces are drawn for each of the blocks. Because the string doesn't stretch, the tension is the same at each end of the string, and the accelerations of the blocks have the same magnitude. Note that we take the positive direction in the direction of the acceleration for each block.

We write SF = ma from the force diagram for each block:

                    y-component (block 1): FT -m1g = m1a;

                    y-component (block 2): m2g - FT = m2a.

By adding the equations, we find the acceleration:

                   a = (m2 - m1)g/(m1+ m2) 

                      = (3.2kg - 2.2kg)(9.80m/s2)/(3.2kg + 2.2kg)

                      = 1.81m/s for both blocks.

For the motion of block 1 we take the origin at the ground and up positive. When block 2 hits the ground, we have

                   v12 = v012 + 2a(y1 - y01)   

                         = 0 + 2(1.81m/s2)(3.60m - 1.80m), which gives v1 = 2.56m/s.

Once block 2 hits the ground,  FT ® 0 and block 1 will have the downward acceleration of g. For this motion of block 1 up to the highest point reached, we have

                   v2 = v12 + 2a(h - y1) 

                   0 = (2.56m/s)2 + 2(-9.80m/s2)(h - 3.60m), which gives h = 3.98m.         

 

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