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Unformatted text preview: 277 Chapter 7 1. (a) The change in kinetic energy for the meteorite would be ( 29( 29 2 2 6 3 14 1 1 4 10 kg 15 10 m/s 5 10 J 2 2 f i i i i K K K K m v ∆ = =  =  =  × × =  × , or 14   5 10 J K ∆ = × . The negative sign indicates that kinetic energy is lost. (b) The energy loss in units of megatons of TNT would be ( 29 14 15 1 megaton TNT 5 10 J 0.1megaton TNT. 4.2 10 J K ∆ = × = × (c) The number of bombs N that the meteorite impact would correspond to is found by noting that megaton = 1000 kilotons and setting up the ratio: 0.1 1000kiloton TNT 8. 13kiloton TNT N × = = 2. With speed v = 11200 m/s, we find 2 5 2 13 1 1 (2.9 10 kg) (11200 m/s) 1.8 10 J. 2 2 K mv = = × = × 3. (a) From Table 21, we have v v a x 2 2 2 = + ∆ . Thus, ( 29 ( 29 ( 29 2 2 7 15 2 7 2 2.4 10 m/s 2 3.6 10 m/s 0.035 m 2.9 10 m/s. v v a x = + ∆ = × + × = × (b) The initial kinetic energy is ( 29( 29 2 2 27 7 13 1 1 1.67 10 kg 2.4 10 m/s 4.8 10 J. 2 2 i K mv = = × × = × The final kinetic energy is ( 29( 29 2 2 27 7 13 1 1 1.67 10 kg 2.9 10 m/s 6.9 10 J. 2 2 f K mv = = × × = × CHAPTER 7 278 The change in kinetic energy is ∆ K = 6.9 × 10 –13 J – 4.8 × 10 –13 J = 2.1 × 10 –13 J. 4. The work done by the applied force a F r is given by cos a a W F d F d φ = ⋅ = r r . From Fig. 724, we see that 25 J W = when φ = and 5.0 cm d = . This yields the magnitude of a F r : 2 25 J 5.0 10 N 0.050 m a W F d = = = × . (a) For 64 φ = ° , we have 2 cos (5.0 10 N)(0.050 m)cos64 11 J. a W F d φ = = × ° = (b) For 147 φ = ° , we have 2 cos (5.0 10 N)(0.050 m)cos147 21 J. a W F d φ = = × ° =  5. We denote the mass of the father as m and his initial speed v i . The initial kinetic energy of the father is K K i = 1 2 son and his final kinetic energy (when his speed is v f = v i + 1.0 m/s) is K K f = son . We use these relations along with Eq. 71 in our solution. (a) We see from the above that K K i f = 1 2 which (with SI units understood) leads to ( 29 2 2 1 1 1 1.0 m/s 2 2 2 i i mv m v = + . The mass cancels and we find a seconddegree equation for v i : 1 2 1 2 2 v v i i = . The positive root (from the quadratic formula) yields v i = 2.4 m/s. (b) From the first relation above K K i = 1 2 son b g , we have 2 2 son 1 1 1 ( /2) 2 2 2 i mv m v = and (after canceling m and one factor of 1/2) are led to v v i son = 2 = 4.8 m s. 279 6. We apply the equation 2 1 2 ( ) x t x v t at = + + , found in Table 21. Since at t = 0 s, x = 0 and 12 m/s v = , the equation becomes (in unit of meters) 2 1 2 ( ) 12 x t t at = + . With 10 m x = when 1.0 s t = , the acceleration is found to be 2 4.0 m/s a =  . The fact that a < implies that the bead is decelerating. Thus, the position is described by 2 ( ) 12 2.0 x t t t = . Differentiating x with respect to t then yields ( ) 12 4.0 dx v t t dt = = ....
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This note was uploaded on 08/09/2010 for the course PHY 202 101A2 taught by Professor Prof.yang during the Summer '10 term at National Taiwan University.
 Summer '10
 Prof.Yang
 Physics, Energy, Kinetic Energy

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