Chapter+8+HW+Solutions

# Chapter+8+HW+Solutions - Chapter 8 HW Solutions 8 We use Eq...

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Chapter 8 HW Solutions 8. We use Eq. 7-12 for W g and Eq. 8-9 for U . (a) The displacement between the initial point and Q has a vertical component of h R downward (same direction as F g ), so (with h = 5 R ) we obtain 2 2 4 4(3.20 10 kg)(9.80 m/s )(0.12 m) 0.15 J g g W F d mgR - = = = × = . (b) The displacement between the initial point and the top of the loop has a vertical component of h – 2 R downward (same direction as F g ), so (with h = 5 R ) we obtain 2 2 3 3(3.20 10 kg)(9.80 m/s )(0.12 m) 0.11 J g g W F d mgR - = = = × = . (c) With y = h = 5 R , at P we find 2 2 5 5(3.20 10 kg)(9.80 m/s )(0.12 m) 0.19 J U mgR - = = × = . (d) With y = R , at Q we have 2 2 (3.20 10 kg)(9.80 m/s )(0.12 m) 0.038 J U mgR - = = × = (e) With y = 2 R , at the top of the loop, we find 2 2 2 2(3.20 10 kg)(9.80 m/s )(0.12 m) 0.075 J U mgR - = = × = (f) The new information ( ) v i 0 is not involved in any of the preceding computations; the above results are unchanged. 34. The distance the marble travels is determined by its initial speed (and the methods of Chapter 4), and the initial speed is determined (using energy conservation) by the original compression of the spring. We denote h as the height of the table, and x as the horizontal distance to the point where the marble lands. Then x = v 0 t and h gt = 1 2 2 (since the vertical component of the marble's “launch velocity” is zero). From these we find x v h g = 0 2 . We note from this that the distance to the landing point is directly proportional to the initial speed. We denote v 0 1 be the initial speed of the first shot and D 1 = (2.20 – 0.27) m = 1.93 m be the horizontal distance to its landing point; similarly, v 02 is the initial speed of the second shot and D = 2.20 m is the horizontal distance to its landing spot. Then

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02 02 01 01 1 1 v D D v v v D D = = When the spring is compressed an amount , the elastic potential energy is 1 2 2 k . When the marble leaves the spring its kinetic energy is 1 2 0 2 mv . Mechanical energy is conserved: 1 2 0 2 1 2 2 mv k = , and we see that the initial speed of the marble is directly proportional to the original compression of the spring. If 1 is the compression for the first shot and 2 is the compression for the second, then v v 02 2 1 01 = b g . Relating this to the previous result,
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