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Unformatted text preview: bello (rtb473) – hw5 – Demkov – (59910) 1 This printout should have 50 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 3) 10.0 points A block of mass 4 . 1 kg, which has an initial velocity of 2 . 4 m / s at time t = 0, slides on a horizontal surface. Calculate the work that must be done on the block to bring it to rest. Correct answer: − 11 . 808 J. Explanation: Let : v = 2 . 4 m / s and v f = 0 . K + W = K f = 0 W = − K = − 1 2 mv 2 = 1 2 (4 . 1 kg) (2 . 4 m / s) 2 = − 11 . 808 J . 002 (part 2 of 3) 10.0 points If a constant friction force of 8 . 1 Newtons is exerted on the block by the surface, what is the acceleration? Correct answer: − 1 . 97561 m / s. Explanation: Let : m = 4 . 1 kg and F = − 8 . 1 N . The force is F = ma a = F m = − 8 . 1 N 4 . 1 kg = − 1 . 97561 m / s 2 . 003 (part 3 of 3) 10.0 points Determine the distance that the block slides as it comes to rest. Correct answer: 1 . 45778 m. Explanation: W = F x x = W F = − 11 . 808 J − 8 . 1 N = 1 . 45778 m . 004 (part 1 of 3) 10.0 points A block is at rest on the incline shown in the figure. The coefficients of static and kinetic friction are μ s = 0 . 6 and μ k = 0 . 51, respec tively. The acceleration of gravity is 9 . 8 m / s 2 . 1 4 k g μ 27 ◦ What is the frictional force acting on the 14 kg mass? Correct answer: 62 . 2875 N. Explanation: F f N mg 27 ◦ The forces acting on the block are shown in the figure. Since the block is at rest, the magnitude of the friction force should be equal to the component of the weight on the plane of the incline F f = M g sin θ = (14 kg) (9 . 8 m / s 2 ) sin 27 ◦ = 62 . 2875 N . 005 (part 2 of 3) 10.0 points What is the largest angle which the incline can have so that the mass does not slide down bello (rtb473) – hw5 – Demkov – (59910) 2 the incline? Correct answer: 30 . 9637 ◦ . Explanation: The largest possible value the static friction force can have is F f,max = μ s N , where the normal force is N = Mg cos θ . Thus, since F f = M g sin θ , M g sin θ m = μ s M g cos θ m tan θ m = μ s θ m = tan − 1 ( μ s ) = tan − 1 (0 . 6) = 30 . 9637 ◦ . 006 (part 3 of 3) 10.0 points What is the acceleration of the block down the incline if the angle of the incline is 38 ◦ ? Correct answer: 2 . 09501 m / s 2 . Explanation: When θ exceeds the value found in part 2, the block starts moving and the friction force is the kinetic friction F k = μ k N = μ k M g cos θ. Newton’s equation for the block then becomes M a = M g sin θ − F f = M g sin θ − μ k M g cos θ and a = g [sin θ − μ k cos θ ] = (9 . 8 m / s 2 ) [sin38 ◦ − (0 . 51) cos 38 ◦ ] = 2 . 09501 m / s 2 ....
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 Spring '07
 Swinney
 mechanics, Force, Friction, Mass, Correct Answer, Bello

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