OH-06-solutions - alsalhi (raa2359) – OH-06 – bradley...

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Unformatted text preview: alsalhi (raa2359) – OH-06 – bradley – (19102) 1 This print-out should have 17 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is Online Homework No. 6 (OH-06). All answers should be submitted online be- fore 2300 on Tuesday 8 March (UAE time). Computers and network connections can give trouble, so do not wait until the last minute! 001 10.0 points A 4 kg object attached to a horizontal string moves with constant speed in a circle on a frictionless horizontal surface. The kinetic energy of the object is 17 J and the tension in the string is 306 N. Find the radius of the circle. Correct answer: 0 . 111111 m. Explanation: Let : m = 4 kg , K = 17 J , and T = 306 N . T N mg r ω The kinetic energy is K = 1 2 mv 2 . The centripital force is supplied by the string: T = F c = mv 2 r r = mv 2 T = 2 K T = 2 (17 J) 306 N = . 111111 m . keywords: 002 10.0 points A 2 . 25 kg block is pushed 1 . 66 m up a vertical wall with constant speed by a constant force of magnitude F applied at an angle of 64 . 5 ◦ with the horizontal. The acceleration of gravity is 9 . 8 m / s 2 . 2 . 25 kg F 6 4 . 5 ◦ If the coefficient of kinetic friction between the block and wall is 0 . 669, find the work done by F . Correct answer: 53 . 7565 J. Explanation: Given : m = 2 . 25 kg , μ = 0 . 669 , θ = 64 . 5 ◦ , and Δ y = 1 . 66 m . F 6 4 . 5 ◦ v mg f k N The block is in equilibrium horizontally, so summationdisplay F x = F cos θ-N = 0 , so that N = F cos θ alsalhi (raa2359) – OH-06 – bradley – (19102) 2 Since the block moves with constant velocity, summationdisplay F y = F sin θ- mg- f k = 0 F sin θ- mg- μ F cos θ = 0 F (sin θ- μ cos θ ) = mg , so F = mg sin θ- μ cos θ = (2 . 25 kg) (9 . 8 m / s 2 ) sin 64 . 5 ◦- . 669 cos 64 . 5 ◦ = 35 . 8785 N . Thus W F = ( F sin θ ) (Δ y ) = (35 . 8785 N) (sin64 . 5 ◦ )(1 . 66 m) = 53 . 7565 J , since 1 J = 1 kg · m 2 / s 2 . 003 10.0 points Starting from rest at a height equal to the radius of the circular track, a block of mass 23 kg slides down a quarter circular track under the influence of gravity with friction present (of coefficient μ ). The radius of the track is 34 m. The acceleration of gravity is 9 . 8 m / s 2 . 34 m 23 kg θ If the kinetic energy of the block at the bottom of the track is 5000 J, what is the work done against friction? Correct answer: 2663 . 6 J. Explanation: W = W f + K W f = mg R- K = (23 kg) (9 . 8 m / s 2 ) (34 m)- (5000 J) = 2663 . 6 J . 004 (part 1 of 2) 10.0 points A 5 . 07 × 10 − 5 kg raindrop falls vertically at constant speed under the influence of gravity and air resistance....
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This note was uploaded on 05/03/2011 for the course PHYS 191 taught by Professor Aboc during the Spring '10 term at The Petroleum Institute.

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OH-06-solutions - alsalhi (raa2359) – OH-06 – bradley...

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