# q2s - Physics 2211 Spring 2008 Quiz#2 Solutions Unless...

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Physics 2211 Quiz #2 Solutions Spring 2008 Unless otherwise directed, all springs, cords, and pulleys are ideal, and drag should be neglected. I . (16 points) A block of mass m accelerates up a frictionless incline that makes an angle θ with the horizontal due to a horizontal applied force of magnitude A . The coefficient of static friction between the block and the ramp is μ s , while the coefficient of kinetic friction between the block and the ramp is μ k . What is the acceleration magnitude of the block, in terms of any or all of m , θ , A , μ s , μ k , and physical or mathematical constants? ( On Earth. ) . . . . . . . . . . . . . . . . . . . . . . . Announced during quiz: the incline is not frictionless. Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration. Resolve any forces that do not lie along an axis into components. Write out Newton’s Second Law for each direction. First, the y direction, from which the normal force, n , will be determined: X F y = n - A y - w y = ma y = 0 n = A sin θ + mg cos θ where A is the magnitude of the applied force and w is the magnitude of the weight force. Next, the x direction: X F x = A x - w x - f k = ma x ma x = A cos θ - mg sin θ - μ k n where f k is the force of kinetic friction. Substituting the expression for the normal force found from the y direction: ma x = A cos θ - mg sin θ - μ k ( A sin θ + mg cos θ ) So a x = A cos θ - mg sin θ - μ k A sin θ - μ k mg cos θ m or a x = A (cos θ - μ k sin θ ) - mg (sin θ + μ k cos θ ) m Quiz #2 Solutions Page 1 of 6

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II . (16 points) A ping-pong ball of mass m is thrown straight downward with a speed that is twice its terminal speed. What is the magnitude of its acceleration upon release, in terms of any or all of m and physical or mathematical constants? ( On Earth, do NOT neglect drag! ) . . . . . . . . . . . . . . . . . . . . . . . Use Newton’s Second Law. Sketch a Free Body Diagram. Choose a coordinate system. In this case, a coordinate system has been chosen so that the x axis points in the direction of the acceleration when the speed is twice the terminal speed. First, we’d like to know how the drag force is related to the weight force. Write out Newton’s Second Law for the situation where the speed, v , is the same as the terminal speed, v t . X F x = D - w = ma x = 0 D = w So 1 4 Av 2 t = mg or 1 2 CρAv 2 t = mg Note that we’ve used D = 1 4 Av 2 , rather than D 1 4 Av 2 . The approximation only means that the factor may be slightly different from 1 4 . In this problem, all that matters is that this factor be a constant, such as 1 2
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