Explanation A skateboarder of mass m, rolls down an incline of length s, with an initial speed, v 0 in a time t. There is no friction between the incline and the skate board. The plane of the incline is at an angle θ to the horizontal. The forces acting on the skateboarder is shown in the diagram below. The weight mg of the skateboarder acts vertically downwards. The normal force F N acts perpendicular to the incline. The weight mg is resolved into two components, m g sin θ and m g cos θ parallel and perpendicular to the incline. The skateboarder accelerates with an acceleration a downwards under the action of the component of weight m g sin θ . F = m g sin θ ……(1) By Newton’s second law, F = m a ……(2) From equations (1)and (2), m a = m g sin θ a = g sin θ ……(3) If the skateboarder rolls down the incline of length s in a time t, use the equation of motion, s = v 0 t + 1 2 a t 2 Using the expression for acceleration found in equation (3)in the above expression and simplifying for sin θ . s = v 0 t + 1 2 a t 2 = v 0 t + 1 2 g sin θ t 2 sin θ = 2 ( s − v 0 t ) g t 2 Given: The initial speed of the skateboarder v 0 = 2...

7th Edition

ISBN: 9780321625915

Author: Douglas C. Giancoli

Publisher: PEARSON

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Chapter 4, Problem 54P

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The angle of inclination of the plane on which a skateboarder rolls down.

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Chapter 4, Problem 53P

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