EAS208_Exam02_MakeUp_102910_FINAL_Solutions

EAS208_Exam02_MakeUp_102910_FINAL_Solutions - EAS208 Fall...

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EAS208 Fall 2010 – Exam 2 Make-up Exam SOLUTIONS Name: Person Number: Problem 1: The block of weight is sliding to the right along the rough, horizontal surface. When the block is at position 40 Wl = b A , its speed is 15 ft v s = . The spring of stiffness 300 lb k ft = brings the block to rest. If the coefficient of kinetic friction between the block and the surface is 0.3 k μ = , determine the maximum force in the spring. Neglect the weight of the end plate that is attached to the spring. (30 points) Solution: The forces that perform work on the block are both conservative and non-conservative. Apply Work-Energy Relation from position A (shown at figure) to position B (maximum deflection of the spring): ' AB g e UV V T = Δ+ Δ () 22 11 0 B A r BA r f dr k S S m v v =+ − + r r r There is no motion in the vertical direction ( y -direction), thus: 0 yy Fm a W N N W Σ= ⇒−=⇒= f Therefore, the friction force is kk f NW = = , 0 A Sf t = , 15 A t v s = , and 0 B ft v s = . Substituting for the above in the Work-Energy Relation: 00 B A r kB r Wd r k S m v −=− + r r r A 6 0 B ft S ft W Wdx kS v g + −= A
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() 22 11 6 kBB W WS k S v g μ −+ = A 21 2 0 kk BB A W W SS W v k g μμ ⎛⎞ ++ ⎜⎟ ⎝⎠ = 2 0.3 40 12 0.3 40 40 15 0 300 300 300 32.2 ⋅⋅ = 2 0.08 0.4517 0 + −= The above is a quadratic equation for . The roots are: B S ,1,2 4 0.08 0.08 4 0.4517 0.08 1.347 B bb a c S a −± ± +⋅ ± == = 2 f t Thus: ,1 0.6335 B S = and ,2 0.7135 B Sf t = −
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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_Exam02_MakeUp_102910_FINAL_Solutions - EAS208 Fall...

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