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05prob_11

# 05prob_11 - Physics 9A-C Assignment#5 Cole...

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Assigment #5, 4/29/11 Page 1 of 3 UC Davis F oe = 30° F N mg f oe = 3 Physics 9A-C Assignment #5 Cole [6.4] Note that the box is pushed along a level floor; the person is pushing down on the box. a) Draw a free-body diagram. Í F y = F N - F sin oe - mg = 0 õ F N = F sin oe + mg. ÍF x = F cos oe - μ k ( F sin oe + mg ) = 0 because the velocity is constant. Solve for F : F = μ k mg cos φ μ k sin φ = 0.25 ( ) 30kg ( ) 9.8 m/s 2 ( ) cos 30¡- 0.25 ( ) sin30 ° = 99.2 N b) F Î x cos oe = (99.2 N)(4.50 m) cos 30° = 386.5 J c) The normal force is F N = mg + F sinoe, and so the work done by friction is W f = –(0.25)[(30 kg)(9.80 m/s 2 ) + (99.2 N) sin 30°](4.50 m) = -386.5 J. d) Both the normal force and gravity act perpendicular to the direction of motion, so neither force does work. e) The net work done is zero (the kinetic energy doesn’t change.) [6.38] W = F dx 0 3 m = area under the curve. a) W = 1 2 (2 m ª 2 N) + (1 m )(2 N) = 4 J. Using the work energy theorem: W = ÎK 1 2 mv 2 - 1 2 mvø 2 õ v = 2 W m = 2 4 J ( ) 2 kg = 2.0 m/s b) No area is accumulated so the speed cannot change. v = 2.0 m/s. c) W = 4 J - 1 2 (2 m)(1 N) = 3 J. v = v = 2 3 J ( ) 2 kg = 1.73 m/s. [6.47] The total power is P = Fv = (165 N)(9.00 m/s) = 1.485 ª 10 3 W, so the power per rider is 742.5 W, or about 1.0 hp (which is a very large output, and cannot be sustained for long periods). [6.68] If this were an ideal spring the applied force to stretch it would be F = kx. The force varies, so we must integrate it: W F = kx bx 2 + cx 3 ( ) dx 0 x = 1 2 kx 2 - 1 3 bx 3 + 1 4 cx 4 = (50.0 N/m ) x 2 2 (233 N/m 2 ) x 2 3 + (3000 N/m 3 ) x 2 4 . a) When x 2 = 0.050 m, W = 0.115 J, or 0.12 J b) When x 2 = - 0.050 m, W = 0.173 J, or 0.17 J. c) It’s easier to stretch the spring; the quadratic bx 2 term is always in the x -direction, and so the needed force, and hence the needed work, will be less when x 2 > 0.

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