Practice Exam 2

# Practice Exam 2 - Version PREVIEW – Practice Exam 2 –...

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Unformatted text preview: Version PREVIEW – Practice Exam 2 – chelikowsky – (59005) 1 This print-out should have 22 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. Elevator Lifted at a Const Speed 001 10.0 points An elevator is being lifted up an elevator shaft at a constant speed by a steel cable as shown in the figure below. All frictional effects are negligible. steel cable Elevator going up at constant speed In this situation, forces on the elevator are such that 1. the upward force by the cable is greater than the sum of the downward force of gravity and a downward force due to the air. 2. the upward force by the cable is greater than the downward force of gravity. 3. the upward force by the cable is equal to the downward force of gravity. correct 4. None of these. (The elevator goes up because the cable is being shortened, not be- cause an upward force is exerted on the eleva- tor by the cable.) 5. the upward force by the cable is smaller than the downward force of gravity. Explanation: Since the elevator is being lifted at a con- stant speed, the net force on it is zero, there- fore, the upward force by the cable is equal to the downward force of gravity. Holt SF 04Rev 21 002 10.0 points A freight train has a mass of 1 . 7 × 10 7 kg. If the locomotive can exert a constant pull of 7 . 2 × 10 5 N, how long would it take to increase the speed of the train from rest to 72 . 5 km / h? (Disregard friction.) 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 475 . 502 s. Explanation: Basic Concepts: vector F net = Σ vector F = mvectora v f = at since v i = 0 m/s. Given: m = 1 . 7 × 10 7 kg F net = 7 . 2 × 10 5 N v f = 72 . 5 km / h Solution: a = F net m = 7 . 2 × 10 5 N 1 . 7 × 10 7 kg = 0 . 0423529 m / s 2 so that Δ t = v f a = 72 . 5 km / h . 0423529 m / s 2 · 1000 m 1 km · 1 h 3600 s = 475 . 502 s Serway CP 04 56 Version PREVIEW – Practice Exam 2 – chelikowsky – (59005) 2 003 10.0 points As a protest against the umpire’s calls, a base- ball pitcher throws a ball straight up into the air at a speed of 24 m / s. In the process, he moves his hand through a distance of 1 . 32 m. The acceleration of gravity is 9 . 8 m / s 2 . If the ball has a mass of 0 . 181 kg, find the force he exerts on the ball to give it this upward speed. 1. @@@ 2. @@@ 3. @@@ 4. @@@ 5. @@@ 6. @@@ 7. @@@ 8. @@@ 9. @@@ 10. @@@ Correct answer: 41 . 2647 N. Explanation: Given : v = 24 m / s , Δ y = 1 . 32 m , and m = 0 . 181 kg . W F The acceleration of the ball is given by v 2 = v 2 i + 2 a Δ y a = v 2 − v 2 i 2 Δ y = v 2 2 Δ y since v i = 0 . From Newton’s second law, summationdisplay F y = F − W = ma F = W + ma = mg + m v 2 2 Δ y = m parenleftbigg g + v 2 2 Δ y parenrightbigg = (0 . 181 kg) bracketleftbigg 9 . 8 m / s 2 + (24 m / s) 2 2 (1 . 32 m) bracketrightbigg = 41 . 2647 N ....
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Practice Exam 2 - Version PREVIEW – Practice Exam 2 –...

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