# HW12Sol - 17.0o mg n fk Jim Guinns PHYS2211 Assignment #12...

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17.0 o mg n f k Jim Guinn’s PHYS2211 Assignment #12 Solutions Due Monday, Mar. 22, 2010 1. A 15.0kg box slides 2.00m down a ramp that is inclined 17.0 o to the horizontal as in the diagram. There is a coefficient of kinetic friction of 0.120 between the block and the ramp. How much work is done by the gravitational force, the normal force, and friction? (Keep in mind that W = F r cos θ .) We have that for the gravitational force, W = F r cos θ = mg r cos(90.0 o – 17.0 o ) = (15.0kg) (9.80m/s 2 ) (2.00m) cos73.0 o = W grav = 86.0J . For the normal force we have W = F r cos θ = F N r cos90.0 o = W norm = 0 . For the frictional force we have W = F r cos θ = μ k N r cos180 o = - μ k mg cos17.0 o r = W = -(0.120) (15.0kg) (9.80m/s 2 ) (cos17.0 o ) (2.00m) = W = -33.7J . 2. A 5.00kg box slides on the floor. The initial speed of the box is 7.50m/s and the coefficient of kinetic friction between the box and the floor is 0.165 . For this problem, use that W

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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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HW12Sol - 17.0o mg n fk Jim Guinns PHYS2211 Assignment #12...

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