HW 6-2-solutions

# HW 6-2-solutions - madrid(tmm2353 – HW 6-2 –...

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Unformatted text preview: madrid (tmm2353) – HW 6-2 – Antoniewicz – (56445) 1 This print-out should have 19 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 2.0 points An object at location (− 17 , , ) m moves to location (− 25 , , ) m. While it is moving it is acted on by a constant force of ( 23 , , ) N. How much work is done on the object by this force? Correct answer: − 184 J. Explanation: W = vector F · Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (23 N )( − 25 − ( − 17))m = − 184 002 (part 2 of 4) 1.0 points Does the kinetic energy of the object increase or decrease? 1. Increase 2. There is no change in the kinetic energy of the ball. 3. Decrease correct Explanation: Consider the update form of the Conserva- tion of Energy: E sys,f = E sys,i + W surr . The system is the object, and the surroundings are doing work on the object. The calculated work due to the force on the object is neg- ative. This means that the initial energy of the system is greater than the final energy of the system, i.e. the energy of the system de- creased. The force acts on the object in a direction that opposes the initial motion of the object. Thus, the kinetic energy of the object decreases. 003 (part 3 of 4) 2.0 points Now, consider a different object as it moves from location (− 25 , , ) m to location (− 17 , , ) m. While it is moving, it is acted on by a constant force of ( 23 , , ) N. How much work is done on the second object by this force? Correct answer: 184 J. Explanation: W = vector F • Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (23 N )( − 17 − ( − 25))m = 184 004 (part 4 of 4) 1.0 points Does the kinetic energy of the second object increase or decrease? 1. There is no change in the kinetic energy of the ball. 2. Increase correct 3. Decrease Explanation: Apply the update form of the Conservation of Energy once again. The work calculated is positive, so the final energy state will have a greater value than the initial energy state. 005 (part 1 of 4) 2.0 points A ball of mass 0 . 8 kg falls downward. Initially, you observe it to be 4 . 25 m above the ground. After a short time, the ball is just about to hit the ground. During this interval, how much work was done on the ball by the gravitational force? Use 9 . 8 m s 2 as the acceleration due to gravity. Correct answer: 33 . 32 J. Explanation: W = vector F · Δ vectorr = F y Δ y madrid (tmm2353) – HW 6-2 – Antoniewicz – (56445) 2 = − mg ( y f − y i ) = − (0 . 8 kg) parenleftbigg 9 . 8 m s 2 parenrightbigg (0 − 4 . 25 m) = 33 . 32 J 006 (part 2 of 4) 1.0 points Does the kinetic energy of the object increase or decrease?...
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HW 6-2-solutions - madrid(tmm2353 – HW 6-2 –...

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