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HW4-phy2048-Spring-2014-solution

# 84 103 j 2 phy2048 spring 2014 hw4 c for

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Unformatted text preview: We use d to denote the magnitude of the spelunker’s displacement during each stage. The mass of the spelunker is m = 80.0 kg. The work done by the lifting force is denoted Wi where i = 1, 2, 3 for the three stages. We apply the work- energy theorem, Eq. 17- 15. 1 (a) For stage 1, W1 − mgd = ΔK1 = 2 mv12 , where v1 = 5.00 m / s. This gives 1 1 W1 = mgd + mv12 = (80.0 kg)(9.80 m/s 2 )(10.0 m) + (80.0 kg)(5.00 m/s)2 = 8.84 ×103 J. 2 2 (b) For stage 2, W2 – mgd = ΔK2 = 0, which leads to W2 = mgd = (80.0 kg)(9.80 m/s 2 )(10.0 m) = 7.84 ×103 J. 2 PHY2048 Spring 2014 HW4 (c) For stage 3, W3 − mgd = ΔK3 = − mv . We obtain 1 1 W3 = mgd − mv12 = (80.0 kg)(9.80 m/s 2 )(10.0 m) − (80.0 kg)(5.00 m/s)2 = 6.84 ×103 J. 2 2 7P24. (a) Using notation common to many vector- capable calculators, we have (from Eq. 7- 8) W = dot([20.0,0] + [0, −(3.00)(9.8)], [0.500 ∠ 30.0º]) = +1.31 J , where “dot” stands for dot product. (b) Eq. 7- 10 (along with Eq. 7- 1) then leads to v = 2(1.31 J)/(3.00 kg) =...
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