Test_2 - Physics 211 Test 2 spring 2008 clarity Some...

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Physics 211 Test 2 spring 2008 Some answers are given with three significant figures for clarity. 1. Body A weighs 211 N, and body B weighs 75 N. The coefficients of friction between A and the incline are µ s = 0.56 and µ k = 0.25. Angle θ is 40°. Find the acceleration of A if A is moving down the incline. Assume the acceleration, a, is in the positive direction in both free-body diagrams. 2. In the figure, blocks A and B have weights of 48 N and 37 N, respectively. (b) Block C suddenly is lifted off A . What is the acceleration of block A , if µ k between A and the table is 0.10? (a) Determine the minimum weight of block C to keep A from sliding, if µ s between A and the table is 0.30. F N - (48 N + W C ) = 0 F T - f SMax = 0 F T – 37 N = 0 f SMax = S F N W C = 75 F N F T F T (48 N +W ) f SMax 37 N a = 0 a = 0 F N F T F T 48 N f k 37 N a a F N - 48 N = 0 F T – f k = M A a 37 N - F T = M B a f k = k F N a = 3.72 m/ M A = 48N/g M B = 37N/g θ +y +x F N F T f k W A F T W B +y W A Sin θ - f k – F T = (W A /g)a and F N - W A Cos θ = 0 and f k = .25F N for A F T – W B = (W B /g)a for B The result of these equations is a = 0.69 m/s down the incline for A v
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3. The figure below shows an initially stationary block of mass m on a floor. A force of magnitude 0.570 mg is then applied at upward angle θ = 21°. What is the magnitude of the acceleration of the block across the floor if s = 0.600 and k = 0.400? FSin θ + F N mg = 0 FCos θ - f = m a Assuming f = . s F N, results in a i indicating that there is motion; thus we must use f = k F N This results in a = 2.1 m/s 2 4. A 2.5 kg block is pushed along a horizontal floor by a force F of magnitude 34 N at an angle θ = 42 o with the horizontal. The coefficient of kinetic friction between the block and floor is 0.25.
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