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Ch6-HW2-solutions

# Ch6-HW2-solutions - louey(cal2859 Ch6-HW2 florin(56930 This...

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louey (cal2859) – Ch6-HW2 – florin – (56930) 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(part1of4)10.0points An object at location (− 20 , 0 , 0 ) m moves to location (− 26 , 0 , 0 ) m. While it is moving it is acted on by a constant force of ( 25 , 0 , 0 ) N. How much work is done on the object by this force? Correct answer: 150 J. Explanation: W = vector F · Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (25 N )( 26 ( 20))m = 150 002(part2of4)10.0points Does the kinetic energy of the object increase or decrease? 1. There is no change in the kinetic energy of the ball. 2. Decrease correct 3. Increase 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(part3of4)10.0points Now, consider a different object as it moves from location (− 26 , 0 , 0 ) m to location (− 20 , 0 , 0 ) m. While it is moving, it is acted on by a constant force of ( 25 , 0 , 0 ) N. How much work is done on the second object by this force? Correct answer: 150 J. Explanation: W = vector F Δ vectorr = F x Δ x + F y Δ y + F z Δ z = (25 N )( 20 ( 26))m = 150 004(part4of4)10.0points Does the kinetic energy of the second object increase or decrease? 1. Decrease 2. Increase correct 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 10.0points 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 0 . 2 m of the wire? Correct answer: 1 × 10 19 J. Explanation: Since the force and displacement are in the same direction, the dot product reduces to a simple scalar product:

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louey (cal2859) – Ch6-HW2 – florin – (56930) 2 W = vector F · Δ vectorr = F Δ r = (5 × 10 19 N)(0 . 2 m) = 1 × 10 19 J .
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Ch6-HW2-solutions - louey(cal2859 Ch6-HW2 florin(56930 This...

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