{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

EAS208_Exam02_MakeUp_102910_FINAL_Solutions

# EAS208_Exam02_MakeUp_102910_FINAL_Solutions - EAS208 Fall...

This preview shows pages 1–3. Sign up to view the full content.

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 W l = 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): ' A B g e U V V T = Δ + Δ + Δ ( ) ( ) 2 2 2 2 1 1 0 2 2 B A r B A B A r f dr k S S m v v = + + r r r There is no motion in the vertical direction ( y -direction), thus: 0 y y F ma W N N W Σ = = = f Therefore, the friction force is k k f N W μ μ = = , 0 A S ft = , 15 A t v s = , and 0 B ft v s = . Substituting for the above in the Work-Energy Relation: ( ) ( ) 2 2 1 1 0 0 2 2 B A r k B r W dr k S m v μ = + r r r A 6 2 2 0 1 1 2 2 B ft S k B ft W W dx kS v g μ + = A

This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document
( ) 2 2 1 1 6 2 2 k B B W W S kS v g μ + = A 2 2 2 12 0 k k B B A W W S S W v k k kg μ μ + + = 2 2 2 0.3 40 12 0.3 40 40 15 0 300 300 300 32.2 B B S S + + = 2 0.08 0.4517 0 B B S S + = The above is a quadratic equation for . The roots are: B S 2 2 ,1,2 4 0.08 0.08 4 0.4517 0.08 1.347 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}