Question.3713 - Homework Assignment for Exam 2 PHY 100- Fall 2024 Chapter 6 Problems due 10/11/24 E1. Assuming force is horizontal ? = 0o wherein cos (0) = 1 Work = F ? d ? cos(?) = 40 x 1.5 x 1 = 60 J. E3. Work = f x d d = work / f = 320 J / 80 N = 4 meters E5. w = Fx ?? d = 50 N x 4 m = 200 J b) wy = 0 J c) wtotal = 200 J E6.w = f x d = 60 x 15 = 900 J The work done on the object tends to be in equilibrium with the increase in kinetic energy = 900 J E9. Spring constant k = 87,600 N/m is compressed by 6.2 cm ?PE = ? kx2 x = 6.2 cm = 0.062 m, ?PE = ? x 87,600 N/m x 0.062 m2 = 0.1684 kJ E10. ?PE = 180 J W = ? x kx2 K = 2w/x2 = 2(180) / 0.302 = 4,000 N/m. E14. PE = mgh PE = 70kg x 9.8m/s2 x 9m = 6,174 J ME = KE + PE = 1,200J + 6,174J = 7,374 J Friction is ignored in this scenario, so there is a potential for mechanical energy to be conserved, converting energy to KEbottom = 7,374 J E15. Mechanical?Energy?at?Point?A = PE + KE = 400 kJ + 130 kJ = 530 kJ Considering work done against friction reduces total mechanical energy, wherein 530 60 KJ = 470 KJ will be Kinetic?Energy?at low?point. Sp1. W = f x d = 6 x 1.7 = 10.2 J Fnet ?= 6 ? 2 = 4N Wnet = 4 x 1.7 = 6.8 J c) Net work done (6.8 J) should be utilized because it accounts for all the forces acting on the block d) because some of the energy is dissipated by heat by friction, accounting for difference between work done by 6N force and net work. e) KE = 6.8 J Velocity = KE = ? ?mv2 V = ?2KE/m = ?2(6.8)/0.38 = 5.98 m/s.

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