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# Chapter-07 - Chapter 7 1(a The change in kinetic energy for...

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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 2-1, we have v v a x 2 0 2 2 = + . Thus, ( 29 ( 29 ( 29 2 2 7 15 2 7 0 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 0 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 - - = = × × = ×

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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. 7-24, we see that 25 J W = when 0 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. 7-1 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 second-degree equation for v i : 1 2 1 2 0 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 0 0 2 ( ) x t x v t at = + + , found in Table 2-1. Since at t = 0 s, x 0 = 0 and 0 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 0 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 = = - . Indeed at t =3.0 s, ( 3.0) 0 v t = = and the bead stops momentarily. The speed at 10 s t = is ( 10) 28 m/s v t = = - , and the corresponding kinetic energy is 2 2 2 1 1 (1.8 10 kg)( 28 m/s) 7.1 J. 2 2 K mv - = = × - = 7. By the work-kinetic energy theorem, ( 29 2 2 2 2 1 1 1 (2.0kg) (6.0m/s) (4.0m/s) 20 J. 2 2 2 f i W K mv mv = ∆ = - = - = We note that the directions of r v f and r v i play no role in the calculation.

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