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Unformatted text preview: obaidi (aao476) – OH-04 – Bradley – (191) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. This is ONLINE HOMEWORK No 4. It is due by 6.00 AM Abu Dhabi Time on Tuesday 9 September (09.00 PM 8 September, Texas Time) unless otherwise notified. Check Quest announcements for hints or updates on home- work assignments. 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. 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.) 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. 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. 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. 002 10.0 points A toy car is given a quick push so that it rolls up an inclined ramp. After it is released, it rolls up, reaches its highest point and rolls back down again. Friction is so small it can be ignored. What net force acts on the car? 1. Net increasing force down the ramp 2. Net force of zero 3. Net constant force up the ramp 4. Net decreasing force down the ramp 5. Net decreasing force up the ramp 6. Net constant force down the ramp cor- rect 7. Net increasing force up the ramp Explanation: The net force is the component of gravity along the incline; it is constant and acts down the ramp. 003 (part 1 of 2) 5.0 points A 23 . 3 kg mass attached to a spring scale rests on a smooth, horizontal surface. The spring scale, attached to the front end of a boxcar, reads T = 33 . 6 N when the car is in motion. m If the spring scale reads zero when the car obaidi (aao476) – OH-04 – Bradley – (191) 2 is at rest, determine the acceleration of the car, when it is in motion as indicated above. Correct answer: 1 . 44206 m / s 2 . Explanation: Basic Concept: Fictitious forces Inertial/non-inertial frames Solution: In the laboratory frame (which is an inertial frame): The block is accelerating with the car, with an acceleration of a . This requires an external force which must equal m a . Since the tension T is the ONLY available, external force we know that m a = T, so that a = T m = 33 . 6 N 23 . 3 kg = 1 . 44206 m / s 2 Alternative derivation in the acceler- ated frame of the car: The acceleration of the mass in the refer- ence frame of the car is a ′ = 0; on the other...
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This note was uploaded on 02/16/2009 for the course PYHS 191 taught by Professor Curtise during the Spring '08 term at Abilene Christian University.
- Spring '08