PHY302K-Sai-HW5-solutions

# PHY302K-Sai-HW5-solutions - Oslund (hno56) – Homework 5...

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Unformatted text preview: Oslund (hno56) – Homework 5 – sai – (58015) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A block of mass m is pushed a horizontal dis- tance D from position A to position B, along a horizontal plane with friction coefficient μ . Then m is pushed from B to A. If the force pushing m from A to B is & P , and the force pushing m from B to A is- & P , what is the total work done by friction? 1. 2 ( μmg- P ) D 2. 3.- 2 μmg D correct 4. +2 μmg D 5. 2 ( P- μmg ) D Explanation: From A to B, the work done by friction is- μmg D . From B to A, the work done by frition is- μmg D too since the work done by kinetic friction is always negative. So the total work done by friction is equal to the sum of these two works, and we get- 2 μmg D . 002 (part 1 of 3) 10.0 points Two students ride in carts opposite to one another in a spinning ferris wheel as shown below. R B A v Student A is originally at the bottom of the ferris wheel while student B is at the top of the ferris wheel. As the wheel turns, student B comes to the bottom while student A arrives at the top, as shown below. R A B v The ferris wheel spins at a constant speed so that the two students are traveling with con- stant speed. Students A and B have masses m A and m B , respectively. The ferris wheel has a radius R . What is the magnitude of the TOTAL work done on student A in moving from the bottom to the top of the ferris wheel? The total work is the sum of the work done by all of the forces on the body; i.e. , W total = F Δ s . Neglect air resistance. 1. W total A = 0 correct Oslund (hno56) – Homework 5 – sai – (58015) 2 2. W total A = 2 ( m B- m A ) g R 3. W wheel B = m A g R 4. W total A = 2 m B g R 5. W wheel B = ( m A + m B ) g R 6. W total A = 2 ( m A- m B ) g R 7. W wheel B = m B g R 8. W total A = 2 m A g R 9. W total A = 2 ( m A + m B ) g R Explanation: The total work involves all the forces acting on a single body; i.e. , W total = & F net · d&s . The work:energy theorem relates the total work done to the change in kinetic energy. The kinetic energy at the top and bottom are the same Δ K = 1 2 mv 2- 1 2 mv 2 = 0, even though the velocity is in opposite directions. Therefore W total = Δ K = 0 . The total work, then, must be zero. Furthermore, the force of gravity is equal in magnitude and opposite to the normal force m&g + & N = 0, and the centripetal force is perpendicular to the motion, so W total = 0 . 003 (part 2 of 3) 10.0 points What is the magnitude of the work done on student B by the ferris wheel in moving from the top to the bottom? 1. W wheel B = 0 2. W wheel B = 2 m B g R correct 3. W wheel B = 2 ( m B- m A ) g R 4. W wheel B = 2 ( m A + m B ) g R 5. W wheel B = m B g R 6. W wheel B = ( m A + m B ) g R 7. W wheel B = m A g R 8. W wheel B = 2 ( m A- m B ) g R 9. W wheel B = 2 m A g R Explanation: W total = 0 = W gravity + W wheel W gravity =...
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## This note was uploaded on 02/22/2012 for the course PHY 302K taught by Professor Kaplunovsky during the Spring '08 term at University of Texas.

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PHY302K-Sai-HW5-solutions - Oslund (hno56) – Homework 5...

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