# HW08Sol - For a constant velocity, we again have a x = 0 ,...

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Jim Guinn’s PHYS2211 Assignment #8 Due Wednesday, Feb. 17, 2010 For problems 1-3, consider the diagram above. A 5kg block is pushed with a force F at an angle of 15 o down from the horizontal as shown. There are coefficients of static and kinetic friction of 0.400 and 0.250 respectively, between the block and the surface. We have that F y = n – W – Fsin15 o = ma y = 0 , so n = W + Fsin15 o . Also F x = Fcos15 o – f = ma x . 1. What is the largest magnitude for F that will allow the block to remain stationary? For the maximum force to remain stationary, we have a x = 0 , so F x = Fcos15 o – f s = ma x = 0 and Fcos15 o = f s = f s,max = μ s n = μ s (W + Fsin15 o ) or F = μ s W / (cos15 o μ s sin15 o ) = F = 22.7N . 2. What is the magnitude of F that will keep the block moving with a constant velocity?

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Unformatted text preview: For a constant velocity, we again have a x = 0 , so F x = Fcos15 o f k = ma x = 0 and 5kg 15 o F s = 0.400 k = 0.250 n W f x y Fcos15 o = f k = k n = k (W + Fsin15 o ) or F = k W / (cos15 o k sin15 o ) = F = 13.6N . 3. What is the magnitude of F that will accelerate the block at 1.00m/s 2 ? With a x = 1.00m/s 2 we have F x = Fcos15 o f k = ma x = (5kg) (1.00m/s 2 ) = 5N , and Fcos15 o = f k + 5N = k n + 5N = k (W + Fsin15 o ) + 5N , or F = ( k W + 5N) / (cos15 o k sin15 o ) = F = 19.1N ....
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## This note was uploaded on 09/25/2011 for the course PHYSICS 22 taught by Professor Lomant during the Spring '11 term at Georgia Perimeter.

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HW08Sol - For a constant velocity, we again have a x = 0 ,...

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