EAS208_Exam01_100810_FINAL_Solutions

EAS208_Exam01_100810_FINAL_Solutions - EAS208 Fall 2010...

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EAS208 Fall 2010 – Exam 1 SOLUTIONS Name: Person Number: Problem 1: A linear spring of stiffness k is to be designed to stop a 20 Mg railroad car traveling at 8 km h within after impact. Find the smallest value of k that will produce the desired result. 400 mm (40 points) Solution: F.B.D. railroad car Units conversion: 20 20,000 mM g k == g 1000 8 2.222 3600 km m h m v s hk m s Equation of motion in x -direction: xx x x k Fm a k x m a a x m Σ= ⇒ −= ⇒ = (1) Equation of motion in y -direction: 0 yy a N W N W m g Σ= ⇒−=⇒== (2) We get no information from Eq. (2). From Eq. (1), we observe that acceleration is a function of position. Thus, apply chain rule and integrate to find the velocity: () 0 0.4 2 00 . 4 2 2.222 0 0 2.222 22 m x m s kd v k k v k ax x v x v d v x d x x md x m m m ⎡⎤ =− ⎢⎥ ⎣⎦ ∫∫ 617,271.6 N k m ⇒= or 5 6.17 10 N k m
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EAS208 Fall 2010 – Exam 1 SOLUTIONS Name: Person Number: Problem 2: The two blocks shown are originally at rest.
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This document was uploaded on 10/27/2011 for the course EAS 208 at SUNY Buffalo.

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EAS208_Exam01_100810_FINAL_Solutions - EAS208 Fall 2010...

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