Page 94 - Schaum's Outline of Theory and Problems of Applied Physics
P. 94

CHAP. 7]                                 ENERGY                                        79



        SOLVED PROBLEM 7.16
              A 128-lb woman skates at a velocity of 15 ft/s. What is her kinetic energy?
                                                     2
                  The woman’s mass is m = w/g = 128 lb/32 ft/s = 4 slugs, and so her kinetic energy is
                                                  1
                                           1  2    	            2
                                      KE = mv =     (4 slugs)(15 ft/s) = 450 ft·lb
                                           2      2
        SOLVED PROBLEM 7.17
              A hammer with a 1.5-kg head is used to drive a nail into a wooden board. If the hammerhead is moving
              at 3 m/s when it strikes the nail and the nail moves 10 mm into the board, find the average force the
              hammerhead exerts on the nail.
                  The kinetic energy of the hammerhead is entirely converted into work done in driving the nail through the
              distance s = 10 mm = 0.01 m. To find the force F exerted on the nail, we proceed as follows:
                                         KE of hammerhead = work done on nail
                                                   1 mv = Fs
                                                      2
                                                   2
                                         mv  2  (1.5kg)(3 m/s) 2
                                     F =    =              = 1350 N = 1.35 kN
                                         2s     (2)(0.01 m)
        SOLVED PROBLEM 7.18
              Find the power output of the engine of a 1200-kg car while the car accelerates from 30 to 100 km/h
              in 10 s.
                  Here
                                               km   1000 m/km

                                       v 1 =  30              = 8.33 m/s
                                               h     3600 s/h
                                               km    1000 m/km

                                       v 2 =  100              = 27.78 m/s
                                                h     3600 s/h
              The work the engine must do to accelerate the car from v 1 to v 2 is equal to the change in the car’s KE, so
                                                  1
                                                     2

                                                                   2
                                                              1
                                                        1
                                                           2
                                  W = KE 2 − KE 1 = mv − mv = m v − v  2
                                                  2  2  2  1  2    2  1
                                       1200 kg        2         2          5
                                     =       [(27.78 m/s) − (8.33 m/s) ] = 4.21 × 10 J
                                         2
              The power needed to provide this amount of work in time t = 10sis
                                                     5
                                         W    4.21 × 10 J
                                                                4
                                     P =    =           = 4.1 × 10 J = 42.1kW
                                          t      10 s
              which is equivalent to
                                                          hp

                                          P = (42.1kW) 1.34   = 56.3hp
                                                          kW
        POTENTIAL ENERGY
        The energy a body has by virtue of its position is called potential energy. A book held above the floor has
        gravitational potential energy because the book can do work on something else as it falls; a nail held near a
        magnet has magnetic potential energy because the nail can do work as it moves toward the magnet; the wound
        spring in a watch has elastic potential energy because the spring can do work as it unwinds.
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