HW 04 solution - gutierrez(ig3472 – HW4 – ditmire...

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Unformatted text preview: gutierrez (ig3472) – HW4 – ditmire – (58216) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A car of mass m , initially at rest at time t = 0, is driven to the right as shown along a straight, horizontal road with the engine caus- ing a constant force vector F to be applied. While moving, the car encounters a resistance force equal to − kvectorv , where vectorv is the velocity of the car and k is a positive constant. Use a dot to represent the center of mass of the car. And on the figure, draw and label vectors to represent all the forces acting on the car as it moves with a velocity vectorv to be the right. car v g O + Determine the horizontal acceleration of the car. 1. a = F + k v m 2. a = F + k m v 3. a = k v m 4. a = F − k m v 5. a = F − k v m correct Explanation: F k v N mg F net = ma. But F net = F − k v , so F − k v = ma a = F − k v m . 002 (part 2 of 2) 10.0 points Derive the equation expressing the velocity of the car as a function of time t . 1. v = F k parenleftBig 1 − e − k t/m parenrightBig correct 2. v = F k parenleftBig 1 + e − k m/t parenrightBig 3. v = F k parenleftBig e − k t/m parenrightBig 4. v = F k parenleftBig 1 − e − k m/t parenrightBig 5. v = F k parenleftBig 1 + e − k t/m parenrightBig Explanation: a = dv dt Using the equation from the previous part dv dt = F − k v m (1) Re-arrange and integrating integraldisplay dv F − k v = integraldisplay 1 m dt (2) Let : u = F − k v du = − k dv − 1 k integraldisplay du u = integraldisplay 1 m dt ln( F − k v ) − ln C = − k m t, where C is a constant. v = 1 k parenleftBig F − Ce − k t/m parenrightBig . To evaluate C , use initial conditions t = 0, v = 0: C = F , so v = F k parenleftBig 1 − e − k t/m parenrightBig gutierrez (ig3472) – HW4 – ditmire – (58216) 2 Equation (2) can also be integrated using limits 0 and v for the left-hand side and 0 and t for the right-hand side to obtain the same answer for full credit. Alternate method to solve equation (1) Recognizing that the solution will be in exponential form, try v = Ae B t + C , where A , B and C are con- stants Substituting into equation (1) AB e B t = F m − k m parenleftBig Ae B t + C parenrightBig AB e B t = parenleftbigg F m − k C m parenrightbigg − k A m e B t Equating coefficients to evaluate B and C , B = − k m ,C = F k Therefore, v = Ae − k t/m + F k To evaluate A , use initial conditions t = 0, v = 0 A = − F k so v = F k parenleftBig 1 − e − k t/m parenrightBig . Sketching a graph of the car’s velocity v as a function of time t , we have t v F k 003 (part 1 of 2) 10.0 points A 2 . 4 kg block rests on a frictionless wedge that has an inclination of 43 ◦ and an acceler- ation to the left such that the block remains stationary relative to the wedge; i.e. , the block does not slide up or down the wedge....
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This note was uploaded on 04/13/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.

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HW 04 solution - gutierrez(ig3472 – HW4 – ditmire...

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