# oldhw 13 - alexander(jra2623 oldhomework 13 Turner(92510 h1...

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alexander (jra2623) – oldhomework 13 – Turner – (92510) 1 This print-out should have 10 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A block starts at rest and slides down a Fric- tionless track except For a small rough area on a horizontal section oF the track (as shown in the fgure below). It leaves the track horizontally, ±ies through the air, and subsequently strikes the ground. The acceleration oF gravity is 9 . 81 m / s 2 . μ =0 . 2 1 . 1 m b b b b b b b b b b b b 484 g h 2 . 2 m 5 . 12 m 9 81 m / s v At what height h above the ground is the block released? Correct answer: 5 . 39891 m. Explanation: Let : x = 5 . 12 m , g = 9 . 81 m / s 2 , m = 484 g , μ = 0 . 2 , = 1 . 1 m , h 2 = - 2 . 2 m , h = h 1 - h 2 , and v x = v . μ b b b b b b b b b b b b m 1 x g v Basic Concepts: Conservation oF Me- chanical Energy U i = U f + K f + W . (1) since v i = 0 m/s. K = 1 2 mv 2 (2) U g = mg h (3) W = μmg ℓ . (4) Choosing the point where the block leaves the track as the origin oF the coordinate system, Δ x = v x Δ t (5) h 2 = - 1 2 g Δ t 2 (6) since a x i = 0 m/s 2 and v y i = 0 m/s. Solution: ²rom energy conservation Eqs. 1, 2, 3, and 4, we have 1 2 mv 2 x = mg ( h - h 2 ) - μmg ℓ v 2 x = 2 g h 1 - 2 μg ℓ h 1 = v 2 x 2 g + μℓ (7) h 2 = - 1 2 g t 2 (6) x = v x t . (5) Using Eq. 6 and substituting t = x v x From Eq. 5, we have h 2 = - 1 2 g p x v x P 2 , so v 2 x = - g x 2 2 h 2 . (8)

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alexander (jra2623) – oldhomework 13 – Turner – (92510) 2 Using Eq. 6 and substituting v 2 x from Eq. 8, we have h 1 = - g x 2 2 h 2 2 g + μℓ = - x 2 4 h 2 + μℓ (9) = - (5 . 12 m) 2 4 ( - 2 . 2 m) + (0 . 2) (1 . 1 m) = 3 . 19891 m , and h = h 1 - h 2 = (3 . 19891 m) - ( - 2 . 2 m) = 5 . 39891 m . 002 (part 1 of 2) 10.0 points A block of mass m is pushed a distance D up an inclined plane by a horizontal force F . The plane is inclined at an angle θ with respect to the horizontal. The block starts from rest and the coeFcient of kinetic friction is μ k . m D μ k F θ If N is the normal force; i.e. , the sum of the ±-component which is perpendicular to the inclined plane and the mg -component which is perpendicular to the inclined plane, what is the work done by friction? 1.
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## This note was uploaded on 10/20/2011 for the course PHY 302K taught by Professor Kaplunovsky during the Summer '08 term at University of Texas.

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oldhw 13 - alexander(jra2623 oldhomework 13 Turner(92510 h1...

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