ch07 - 1 With speed v = 11200 m/s we find K= 1 2 1 mv =(2.9...

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1. With speed v = 11200 m/s, we find K mv = = × = × 1 2 1 2 2 9 10 11200 18 10 2 5 2 13 ( . ) ( ) . J.
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2. (a) The change in kinetic energy for the meteorite would be ( )( ) 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 ( ) 14 15 1 megaton TNT 5 10 J 0.1megaton TNT. 4.2 10 J K § · −∆ = × = ¨ ¸ × © ¹ (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 × = =
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3. (a) From Table 2-1, we have v v a x 2 0 2 2 = + . Thus, ( ) ( ) ( ) 2 2 7 15 7 0 2 2.4 10 2 3.6 10 0.035 2.9 10 m/s. v v a x = + = × + × = × (b) The initial kinetic energy is ( )( ) 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 ( )( ) 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 = = × × = × The change in kinetic energy is K = (6.9 × 10 –13 – 4.8 × 10 –13 ) J = 2.1 × 10 –13 J.
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4. 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 = = × =
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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 ( ) 2 2 1 1 1 1.0 2 2 2 i i mv m v ª º = + « » ¬ ¼ . 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 1 2 1 2 1 2 2 2 mv m v i = F H G I K J F H G I K J son 2 and (after canceling m and one factor of 1/2) are led to v v i son = 2 = 4.8 m s.
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6. By the work-kinetic energy theorem, ( ) 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 & v f and & v i play no role in the calculation.
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7. Eq. 7-8 readily yields (with SI units understood) W = F x x + F y y = 2cos(100º)(3.0) + 2sin(100º)(4.0) = 6.8 J.
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8. Using Eq. 7-8 (and Eq. 3-23), we find the work done by the water on the ice block: ( ) ( ) 3 ˆ ˆ ˆ ˆ 210i 150 j 15i 12j (210)(15) ( 150) ( 12) 5.0 10 J. W F d = = = + − = × & &
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9. Since this involves constant-acceleration motion, we can apply the equations of Table 2-1, such as x v t at = + 0 1 2 2 (where x 0 0 = ). We choose to analyze the third and fifth points, obtaining 2 0 2 0 1 0.2m (1.0 s) (1.0 s) 2 1 0.8m (2.0 s) (2.0 s) 2 v a v a = + = + Simultaneous solution of the equations leads to 0 0 v = and a = 0 40 . m s 2 . We now have two ways to finish the problem. One is to compute force from
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