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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.
- Spring '11