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Unformatted text preview: fryer (vdf96) – Conservation of Mechanical Energy – graves – (6) 1 This printout should have 21 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A 0 . 9 kg rock slides horizontally off a table from a height of 1 . 3 m. The speed of the rock as it leaves the thrower’s hand at the edge of the table is 2 . 1 m / s, as shown. The acceleration of gravity is 9 . 81 m / s 2 . 2 . 1 m / s 1 . 3m Δ x How much time does it take the rock to travel from the edge of the table to the floor? Correct answer: 0 . 514816 s. Explanation: Let : h = 1 . 3 m and g = 9 . 81 m / s 2 . The motion of the rock in the ydirection is a free falling motion from rest: h = 1 2 g t 2 t = radicalBigg 2 h g = radicalBigg 2 (1 . 3 m) 9 . 81 m / s 2 = . 514816 s . 002 (part 2 of 2) 10.0 points What is the kinetic energy of the rock just before it hits the floor? Correct answer: 13 . 4622 J. Explanation: Let : m = 0 . 9 kg and v = 2 . 1 m / s . By the workenergy theorem, K f K i = W , where in the present problem, W = ( mg ) h and K i = 1 2 mv 2 . Thus, K f = K i + W = 1 2 mv 2 + mg h = m parenleftbigg 1 2 v 2 + g h parenrightbigg = (0 . 9 kg) bracketleftbigg 1 2 (2 . 1 m / s) 2 +(9 . 81 m / s 2 ) (1 . 3 m) bracketrightbigg = 13 . 4622 J . 003 (part 1 of 6) 10.0 points A projectile of mass 0 . 224 kg is shot from a cannon, at height 6 . 4 m, as shown in the figure, with an initial velocity v i having a horizontal component of 6 . 5 m / s. The projectile rises to a maximum height of Δ y above the end of the cannon’s barrel and strikes the ground a horizontal distance Δ x past the end of the cannon’s barrel. Δ x v i 4 4 ◦ Δ y 6 . 4m fryer (vdf96) – Conservation of Mechanical Energy – graves – (6) 2 Determine the vertical component of the initial velocity at the end of the cannon’s bar rel, where the projectile begins its trajectory. The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 6 . 27698 m / s. Explanation: Let : v xi = 6 . 5 m / s and θ = 44 ◦ . tan θ = v yi v xi v yi = v xi tan θ = (6 . 5 m / s) tan44 ◦ = 6 . 27698 m / s . 004 (part 2 of 6) 10.0 points Determine the maximum height Δ y the pro jectile achieves after leaving the end of the cannon’s barrel. Correct answer: 2 . 01023 m. Explanation: Let : v y top = 0 . v 2 yi = v 2 y top + 2 g Δ y = 2 g Δ y Δ y = radicalBigg v 2 yi 2 g = radicalBigg (6 . 27698 m / s) 2 2 (9 . 8 m / s 2 ) = 2 . 01023 m . 005 (part 3 of 6) 10.0 points Find the work done by the gravitational force on the projectile during the motion in its trajectory. Correct answer: 14 . 0493 J. Explanation: Let : ( x i , y i ) = (0 m , 6 . 4 m) ....
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This note was uploaded on 12/11/2011 for the course PHYSICS 101 taught by Professor Graves during the Fall '10 term at San Jose City College.
 Fall '10
 Graves
 Physics, Energy, Kinetic Energy

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