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212samplefinalsoln - 212 Sample Final exam Solution Dec...

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212 Sample Final exam Solution Dec, 2004 1) A truck applies a torque to its wheels resulting in a forward force on the truck of 500 pounds. The truck itself (without the crate) has a mass of 70 slugs. A 30 slug crate sits in the back of the truck on a horizontal surface with coefficient of friction μ = 0.1. As the truck accelerates from rest, the crate slides backwards relative to the truck. The free body diagrams for this system are: a) Solve for f, and for a crate and a truck relative to the ground If the crate starts a distance of 10 feet from the back of the truck, b) How long will it be before the crate collides with the back wall? c) What speed will the truck be going right before the collision? d) What velocity (speed and direction!) will the crate have relative to the ground right before collision? e) If the crate sticks to the back wall, what speed will the composite body have right after the collision? Neglect the rotary inertia of the wheels. a) f = m crate a crate , but f = μ N = μ m c g, so a crate = μ g = 3.22 ft/sec 2 to the right! 500 - f = m truck a truck , so a truck = 5.76ft/sec^2 to the right b) -10 feet = (1/2) a rel t 2 , with a rel = -2.54 , i.e., 2.54 to the left. so t = 2.81 seconds c) a truck x 2.81sec = 16.19 ft/sec to the right d) a crate x 2.81 = 9.05 ft/sec to the right e) m T 16.19 + m c 9.05 = L total = 1404 slug ft / sec NB this is also 500 lbf times 2.81 seconds ! = (m T +m c ) v final . So v final = 14.04 ft/second 2) An ice skater, of mass 70 kg, going a speed of 6 m/sec, collides with another ice skater (mass 50 kg) who is initially at rest. They entangle and fall to the ice, and then slide with a coefficient of friction of 0.12 How much time elapses until the composite comes to rest? What distance do they slide? Analyze collision: 70 kg * 6m/sec + 50 kg * 0m/sec = 420 kg m/sec = L initial = L final = 120 kg * vf so vf = 420 / 120 = 3.5 m/sec = speed of composite after collision Analyze sliding using impulse momentum: L initial = 420 kg m/sec = Lfinal(=0) - f t, where f = μ Mg = (.12) (120) (9.81) = 141.3 so t = 420/141.3= 2.97 seconds Analyze sliding using work-energy
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