HW_Solns_6

# HW_Solns_6 - Physics 112 HW#6 Solutions 6-4[Sliding Crate...

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- 1 - Physics 112 - HW #6 Solutions Spring 2006 6-4 [Sliding Crate] (a) Newton’s second law and our knowledge of friction give: N = F w , F f = F w , and F f = μ N . Combining these equations yields: F w = μ mg = (0.25)(30.0kg)(9.80m/s 2 ) = 73.5N (b) W w = F w •s = (73.5 N)(4.5 m) = 331 J (c) Since F f = - F w , W f = - W w = -331 J (d) Both of these are perpendicular to the direction of motion so W N = W g = 0 . (e) The net work is zero . 6-9 [Ball on String] (a) W T = T d s = 0 because the tension is always perpendicular to the path of motion. W g = - mg Δ y = 0 because Δ y = 0 for the entire trip, OR . . . F g d s = F g d s = 0 because the net displacement is zero. (b) W T = T d s = 0 because the tension is always perpendicular to the path, while W g = F g • s = - |F g |(2R) = - (0.0800 kg)(9.8 N/kg)(2)(1.60 m) = - 25.1 J The answer is negative because the force of gravity acts in the opposite direction to the displacement. 6-24 [Watermelon Work] (a) Work done by gravity is W = F • s = mg s = (4.80 kg)(9.81 m/s 2 )(25.0 m) = 1180 J (b) Increase in KE of melon is equal to the work done be gravity. Initially melon was stationary, so K f = W = 1180 J (c) Final KE is K f = 1 2 mv f 2 , so final speed is v f = 2K

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HW_Solns_6 - Physics 112 HW#6 Solutions 6-4[Sliding Crate...

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