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# test 2 - Overton Mays Midterm 2 Due Mar 9 2005 10:00 pm...

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Overton, Mays – Midterm 2 – Due: Mar 9 2005, 10:00 pm – Inst: Turner 1 This print-out should have 18 questions. Multiple-choice 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 57 Δ y 6 . 9 m 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) cos 57 = 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 = m g y i . From Δ K = 1 2 m v 2 f - 1 2 m v 2 i and the work-energy theorem Δ K = W , the final velocity is v f = r 2 W m + v 2 i = r 2 m g y i m + v 2 i = q 2 g y i + v 2 i

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Overton, Mays – 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) tan 57 = 9 . 70115 m / s , and the vertical velocity v y top = 0 at the top, 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 . 003 (part 1 of 1) 10 points Betty weighs 419 N . She is sitting on a play- ground swing seat that hangs 0 . 48 m above the ground in its rest position. Her initial speed is zero and her initial height above the ground is 1 . 69 m . At some later time her speed is 1 . 23 m / s and her height above the ground is 1 m .
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test 2 - Overton Mays Midterm 2 Due Mar 9 2005 10:00 pm...

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