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Unformatted text preview: 1 Chapter 7 1. The kinetic energy is given by K = 1 2 mv 2 , where m is the mass and v is the speed of the electron. The speed is therefore v = s 2 K m = s 2(6 . 7 10 19 J) 9 . 11 10 31 kg = 1 . 2 10 6 m / s . 2. (a) The change in kinetic energy for the meteorite would be K = K f K i = K i = 1 2 m i v 2 i = 1 2 4 10 6 kg 15 10 3 m / s 2 = 5 10 14 J where the negative sign indicates that kinetic energy is lost. (b) The energy loss in units of megatons of TNT would be K = 5 10 14 J 1 megaton TNT 4 . 2 10 15 J = 0 . 1 megaton TNT . (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: N = . 1 1000 kiloton TNT 13 kiloton TNT = 8 . 3. We convert to SI units (where necessary) and use K = 1 2 mv 2 . (a) K = 1 2 (110)(8 . 1) 2 = 3 . 6 10 3 J. (b) Since 1000 g = kg, we find K = 1 2 4 . 2 10 3 kg (950 m / s) 2 = 1 . 9 10 3 J . (c) We note that the conversion from knots to m/s can be obtained from the information in Appendix D (knot = 1 . 688 ft/s where ft = 0 . 3048 m), which is also where the ton kilogram conversion can be found. Therefore, K = 1 2 91400 tons) 907 . 2 kg ton ! (32 knots) . 515 m / s knot ! 2 = 1 . 1 10 10 J . 4. We denote the mass of the father as m and his initial speed v i . The initial kinetic energy of the father is K i = 1 2 K son and his final kinetic energy (when his speed is v f = v i + 1 . 0 m/s) is K f = K son . We use these relations along with Eq. 71 in our solution. (a) We see from the above that K i = 1 2 K f which (with SI units understood) leads to 1 2 mv 2 i = 1 2 1 2 m ( v i + 1 . 0) 2 . The mass cancels and we find a seconddegree equation for v i : 1 2 v 2 i v i 1 2 = 0 . The positive root (from the quadratic formula) yields v i = 2 . 4 m/s. (b) From the first relation above ( K i = 1 2 K son ), we have 1 2 mv 2 i = 1 2 1 2 m 2 v 2 son and (after canceling m and one factor of 1 / 2) are led to v son = 2 v i = 4 . 8 m/s. 5. (a) From Table 21, we have v 2 = v 2 + 2 a x . Thus, v = q v 2 + 2 a x = q (2 . 4 10 7 ) 2 + 2 (3 . 6 10 15 ) (0 . 035) = 2 . 9 10 7 m / s . (b) The initial kinetic energy is K i = 1 2 mv 2 = 1 2 1 . 67 10 27 kg 2 . 4 10 7 m / s 2 = 4 . 8 10 13 J . The final kinetic energy is K f = 1 2 mv 2 = 1 2 1 . 67 10 27 kg 2 . 9 10 7 m / s 2 = 6 . 9 10 13 J . The change in kinetic energy is K = 6 . 9 10 13 4 . 8 10 13 = 2 . 1 10 13 J. 6. Using Eq. 78 (and Eq. 323), we find the work done by the water on the ice block: W = ~ F ~ d = 210  150 j 15  12 j = (210)(15) + ( 150)( 12) = 5 . 10 3 J ....
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This note was uploaded on 11/12/2011 for the course PHYS 2001 taught by Professor Sprunger during the Fall '08 term at LSU.
 Fall '08
 SPRUNGER
 Physics, Energy, Kinetic Energy, Mass

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