This preview shows pages 1–3. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.View Full Document
Unformatted text preview: ouseph (jao888) HW03 tsoi (92940) 1 This print-out should have 24 questions. Multiple-choice questions may continue on the next column or page find all choices before answering. 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 downward force of gravity. 2. the upward force by the cable is smaller than the downward force of gravity. 3. 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. 4. the upward force by the cable is equal to the downward force of gravity. correct 5. 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.) 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 Is it correct to say that no force acts on a body at rest? 1. All are wrong. 2. No net force acts on a body at rest; when the net force is zero, the body is in static equilibrium. correct 3. No force acts on a body at rest; if at least one force acted on it the body would move. 4. No force acts on a body at rest; all forces cancel each other. 5. No net force acts on a body at rest; no force acts on the body at all. Explanation: There may be any number of forces that act to produce a zero net force. When the net force is zero, the body is in static equilibrium. 003 10.0 points An elevator starts from rest with a constant upward acceleration and moves 1 m in the first 1 . 6 s. A passenger in the elevator is holding a 7 . 4 kg bundle at the end of a vertical cord. What is the tension in the cord as the ele- vator accelerates? The acceleration of gravity is 9 . 8 m / s 2 . Correct answer: 78 . 3013 N. Explanation: h = 1 m , t = 1 . 6 s , m = 7 . 4 kg , and g = 9 . 8 m / s 2 . T mg a elevator g ouseph (jao888) HW03 tsoi (92940) 2 Let h be the distance traveled and a the acceleration of the elevator. Since the initial velocity is zero, h = v t + 1 2 a t 2 = 1 2 a t 2 a = 2 h t 2 . The equation describing the forces acting on the bundle is F net = ma = T mg T = m ( g + a ) = m parenleftbigg g + 2 h t 2 parenrightbigg = (7 . 4 kg) bracketleftbigg 9 . 8 m / s 2 + 2 (1 m) (1 . 6 s) 2 bracketrightbigg = 78 . 3013 N . 004 (part 1 of 3) 10.0 points You are driving a car down a straight road at a constant 60 miles per hour....
View Full Document