F03s(3) - Physics 4A FINAL EXAM Chapters 1 15 Fall 2003...

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FINAL EXAM Chapters 1 - 15 Fall 2003 1 3m/s 5.5s mg F n F f Name:_________________________________________ Posting Code _____ _____ _____ Solve the following problems in the space provided. Use the back of the page if needed. Each problem is worth 10 points. You must show your work in a logical fashion starting with the correctly applied physical principles which are on the last page. Your score will be maximized if your work is easy to follow because partial credit will be awarded. 1. A coin slides off a 1.20m high horizontal counter and strikes the floor 45.0cm from the base of the counter. Find (a)the time the coin is in the air and (b)the speed that the coin left the counter. (a)The time can be found using the kinematic equation for y, y = y o + v oy t + 1 2 a y t 2 y o =- 1 2 a y t 2 t = - 2 y o a y . Putting in the numbers, t = - 2(1.20) - 9.80 t = 0.495 s . (b)Using the kinematic equation for x to solve for the initial speed, x = x o + v ox t + 1 2 a x t 2 x = v ox t v ox = x t . Putting in the numbers, v ox = 0.450 0.495 v ox = 0.909 m / s . 2. A cart of mass 350g is given a push in front of a motion detector. The resulting velocity versus time graph for the cart is shown. Find (a)the frictional force and (b)the coefficient of kinetic friction. (a)The acceleration can be found by estimating the slope, a v t 3 5.5 a 0.545 m / s 2 . Since the only horizontal force acting on the cart is friction, the Second Law requires, Σ F x = ma x F f = ma = (0.350)(0.545) F f = 0.191 N . (b)Applying the Second Law to the vertical forces, Σ F y = ma y F n - mg = 0 F n = mg . The definition of the coefficient of friction is, F f F n = F f F n = ma mg = a g = 0.545 9.80 = 0.0556 . h = 1.20m
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F03s(3) - Physics 4A FINAL EXAM Chapters 1 15 Fall 2003...

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