Chapter07

# Chapter07 - 1. (a) The change in kinetic energy for the...

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1. (a) The change in kinetic energy for the meteorite would be () ( ) 2 26 3 1 4 11 41 0k g1 51 0m / s 0J 22 fi i i i KK K K m v Δ= − = − = = × × = −× , or 14 || 5 1 0 J K Δ=× . 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 §· −Δ = × = ¨¸ × ©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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2. With speed v = 11200 m/s, we find 25 2 1 3 11 (2.9 10 kg) (11200 m/s) 1.8 10 J. 22 Km v ==× = ×
3. (a) From Table 2-1, we have vv a x 2 0 2 2 =+Δ . Thus, () ( ) 2 27 1 5 2 7 0 2 2.4 10 m/s 2 3.6 10 m/s 0.035 m 2.9 10 m/s. a x =+ Δ = × + × = × (b) The initial kinetic energy is ( ) 2 22 7 7 1 3 0 11 1.67 10 kg 2.4 10 m/s 4.8 10 J. i Km v −− == × × = × The final kinetic energy is ( ) 2 7 7 1 3 1.67 10 kg 2.9 10 m/s 6.9 10 J. f v × × = × The change in kinetic energy is Δ K = 6.9 × 10 –13 J – 4.8 × 10 –13 J = 2.1 × 10 –13 J.

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4. The work done by the applied force a F G is given by cos aa WFdF d φ =⋅ = G G . From Fig. 7-24, we see that 25 J W = when 0 = and 5.0 cm d = . This yields the magnitude of a F G : 2 5.0 10 N 0.050 m a W F d == = × . (a) For 64 , we have 2 cos (5.0 10 N)(0.050 m)cos64 11 J. a WF d × ° = (b) For 147 , we have 2 cos (5.0 10 N)(0.050 m)cos147 21 J. a d × ° =
5. We denote the mass of the father as m and his initial speed v i . The initial kinetic energy of the father is KK i = 1 2 son and his final kinetic energy (when his speed is v f = v i + 1.0 m/s) is f = son . We use these relations along with Eq. 7-1 in our solution. (a) We see from the above that if = 1 2 which (with SI units understood) leads to () 2 2 11 1 1.0 m/s 22 2 ii mv m v ªº =+ «» ¬¼ . The mass cancels and we find a second-degree equation for v i : 1 2 1 2 0 2 vv −−= . The positive root (from the quadratic formula) yields v i = 2.4 m/s. (b) From the first relation above i = 1 2 son bg , we have son 1 (/ 2 ) 2 i mv m v §· = ¨¸ ©¹ and (after canceling m and one factor of 1/2) are led to i son =2 =4.8 ms.

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6. We apply the equation 2 1 00 2 () x tx v ta t =+ + , 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 tt a t . 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 ttt . Differentiating x with respect to t then yields 12 4 dx vt t dt == . Indeed at t =3.0 s, ( 3.0) 0 and the bead stops momentarily. The speed at 10 s t = is (1 0 ) 2 8 m / s , and the corresponding kinetic energy is 22 2 11 (1.8 10 kg)( 28 m/s) 7.1 J. Km v × =
7. By the work-kinetic energy theorem, () 22 2 2 111 (2.0kg) (6.0m/s) (4.0m/s) 20 J. 222 fi W K mv mv =Δ = = = We note that the directions of G v f and G v i play no role in the calculation.

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8. Eq. 7-8 readily yields W = F x Δ x + F y Δ y =(2.0 N)cos(100º)(3.0 m) + (2.0 N)sin(100º)(4.0 m) = 6.8 J.