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**Unformatted text preview: **chauhan (mmc2762) – Ch6-H2 – turner – (56910) 1 This print-out should have 18 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 4) 10.0 points An object at location (− 16 , , ) m moves to location (− 28 , , ) m. While it is moving it is acted on by a constant force of ( 25 , , ) N. How much work is done on the object by this force? Correct answer: − 300 J. Explanation: W = vector F · Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (25 N )( − 28 − ( − 16))m = − 300 002 (part 2 of 4) 10.0 points Does the kinetic energy of the object increase or decrease? 1. Increase 2. Decrease correct 3. There is no change in the kinetic energy of the ball. 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) 10.0 points Now, consider a different object as it moves from location (− 28 , , ) m to location (− 16 , , ) m. While it is moving, it is acted on by a constant force of ( 25 , , ) N. How much work is done on the second object by this force? Correct answer: 300 J. Explanation: W = vector F • Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (25 N )( − 16 − ( − 28))m = 300 004 (part 4 of 4) 10.0 points Does the kinetic energy of the second object increase or decrease? 1. Increase correct 2. Decrease 3. There is no change in the kinetic energy of the ball. 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 10.0 points An electron traveling through a curving wire in an electric circuit experiences a constant force of 5 × 10 − 19 N, always in the direction of its motion through the wire. How much work is done on the electron by this force as it travels through 1 . 2 m of the wire? Correct answer: 6 × 10 − 19 J. Explanation: Since the force and displacement are in the same direction, the dot product reduces to a simple scalar product: chauhan (mmc2762) – Ch6-H2 – turner – (56910) 2 W = vector F · Δ vectorr = F Δ r = (5 × 10 − 19 N)(1 . 2 m) = 6 × 10 − 19 J ....

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