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Unformatted text preview: Trevino, Denise – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner 1 This printout should have 18 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. The due time is Central time. 001 (part 1 of 1) 10 points An elevator has a mass of 1100 kg and car ries a maximum load of 777 kg . A constant frictional force of 4600 N retards its motion upward. The acceleration of gravity is 9 . 8 m / s 2 . What power must the motor deliver at a instantaneous speed of 3 . 61 m / s if the elevator is designed to provide an upward acceleration of 1 . 16 m / s 2 ? Correct answer: 90870 . 6 W. Explanation: The motor must supply the force ~ T that pulls the elevator upward. Applying Newton’s second law to the elevator gives T f M g = M a. Thus T = M ( a + g ) + f = (1877 kg)(1 . 16 m / s 2 + 9 . 8 m / s 2 ) + 4600 N = 25171 . 9 N . and the required instantaneous power is P = T v = (25171 . 9 N)(3 . 61 m / s) = 90870 . 6 W . 002 (part 1 of 1) 10 points A projectile of mass 0 . 596 kg is shot from a cannon, at height 6 . 9 m, as shown in the figure, with an initial velocity v i having a horizontal component of 6 . 3 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. The acceleration of gravity is 9 . 8 m / s 2 . Δ x v i 5 7 ◦ Δ y 6 . 9m Find the magnitude of the final velocity vector when the projectile hits the ground. Correct answer: 16 . 4025 m / s. Explanation: Let : v x i = 6 . 3 m / s , x i = 0 m , y i = 6 . 9 m , and θ = 57 ◦ . For the horizontal component of vector v i , we have v x i = v i cos θ ; consequently v i = v x i cos θ = (6 . 3 m / s) cos57 ◦ = 11 . 5673 m / s . The work done by gravity depends only on the vertical distance y i , since the work involving Δ y adds and subtracts; i.e. , cancels. The work done by gravity is W = mg y i . From Δ K = 1 2 mv 2 f 1 2 mv 2 i and the workenergy theorem Δ K = W , the final velocity is v f = r 2 W m + v 2 i = r 2 mg y i m + v 2 i = q 2 g y i + v 2 i Trevino, Denise – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner 2 = h 2(9 . 8 m / s 2 )(6 . 9 m) + (11 . 5673 m / s) 2 i 1 / 2 = 16 . 4025 m / s . Alternate Solution: The initial velocity in the vertical direction is v y i = v x i tan θ = (6 . 3 m / s) tan57 ◦ = 9 . 70115 m / s , and theverticalvelocity v y top = 0atthetop, there fore we have v 2 y i = v 2 y top + 2 g Δ y , so Δ y = s v 2 y i 2 g = s (9 . 70115 m / s) 2 2(9 . 8 m / s 2 ) = 4 . 80165 m . The magnitude of the final vertical velocity v y f is  v y f  = p 2 g [ y i + Δ y ] = q 2(9 . 8 m / s 2 )[(6 . 9 m) + (4 . 80165 m)] = 15 . 1444 m / s , so v f = q v 2 y f + v 2 x i = q (15 . 1444 m / s) 2 + (6 . 3 m / s) 2 = 16 . 4025 m / s ....
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This note was uploaded on 02/09/2010 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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