SampleExam_2_09s - Sample Exam 2 Short answer 1 You whirl a...

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Sample Exam 2 Short answer. 1. You whirl a rock on a string in a vertical circle. What is the minimum speed that the rock needs to have at the top so that the string does not become limp? Note: You do not need the mass of the rock. mv 2 r = T + mg T = 0 mv 2 r = mg v = rg 2. You are rescued by rope. The rope pulls you upward with a constant acceleration of 0.5 m/s2. Your mass 80 kg. What is the tension in the rope. ma = T mg T = ma + mg = 80 kg 0.5 m / s 2 + 80 kg 9.8 m / s 2 = 824 N 3. A mass of 10 kg is acted on by two forces: F1 = 2 i + 4 j and F2 = 5i -3 j. Find the net force and the acceleration in i, j, k components. F net = F 1 + F 2 = 7 ˆ i + 1 ˆ j a = F net m = 7 10 ˆ i + 1 10 ˆ j
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4. A frictional force with coefficient m acts on a car’s tires. Derive an expression for the minimum radius curve the car can negotiate if it has velocity v 0 and there is no banking. mv 2 r = μ N N = mg mv 2 r = μ mg r = v 2 μ g 5. A mass falls through air with constant velocity. If the m=100 kg, what must be the frictional force acting on it? If the mass falls with constant velocity, the frictional force must balance against the weight to produce zero net force. F f = mg = 100 kg 9.8 m / s 2 = 980 N
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6. A 50kg mass slides down an incline. The incline makes an angle of 30 degrees with horizontal and the mass slides a distance of 100 m. What is its velocity at the bottom of the incline? θ mg N mg N θ x y We begin by computing the acceleration due to the forces that we have drawm. 0 = N mg cos θ N = mg cos θ m a = mg sin θ ma = mg sin θ = mg sin θ = mg sin θ a = g sin θ x i = 0 x f = d v i = 0 v f = ? a = g sin θ v f 2 = v i 2 + 2 a ( x f x i ) v f 2 = 0 + 2 g sin θ ( d 0) v f = 2 g sin θ d = 31.3 m / s 7. A projectile is fired at an angle of 60 degrees and it rises to a maximum height of 100 m. What was the initial magnitude of the velocity of the projectile? When will it land? Where will it land? First we solve for the initial velocity... y f = 100 m y i = 0 v iy = v 0 sin60 v fy = 0 a y = g v fy 2 = v iy 2 + 2 a ( y f y i ) 0 = v iy 2 2 g ( y f 0) v iy 2 = 2 gy f v iy = 2 gy f = 2 9.8 m / s 2 100 m = 44.3 m / s v 0 = v iy sin60 = 51.1 m / s Note: This is another way to do it--different from the way that I did it in class.
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Now compute the total time of flight to land. (Note:
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