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Unformatted text preview: homework 15 – BAUTISTA, ALDO – Due: Feb 27 2006, 4:00 am 1 Version number encoded for clicker entry: V1:1, V2:4, V3:2, V4:4, V5:3. Question 1 Part 1 of 3. 10 points. Assume: The pulleys are weightless, the rope does not stretch, and the system moves at a constant speed which is slow enough that the kinetic energy is negligible. A weight of 738 N is raised by a twopulley arrangement as shown in the figure. m d F Δ x Note: Figure may not be drawn to scale. How much work is done by the agent (force ~ F ) to raise the weight by a vertical distance of 21 m? Correct answer: 15498 J (tolerance ± 1 %). Explanation: Basic Concepts: W = ~ F · ~s U grav . = m g h ~ F net = X i ~ F i Solution: The work done by the force F results in the change of potential energy of the system. The final energy is the potential energy P E final = m g Δ x ; the initial energy is zero. Thus Δ E = m g Δ x = (738 N) (21 m) = 15498 J . Question 2 Part 2 of 3. 10 points. Which of the following is the SI unit of the force [ F ]? 1. [ F ] = W 2. [ F ] = J 3. [ F ] = m/kg · s 4. [ F ] = m/s 2 5. [ F ] = J s 6. [ F ] = kg · m/s 7. [ F ] = N/s 8. [ F ] = kg 9. [ F ] = kg · m/s 2 correct 10. [ F ] = N m Explanation: The SI unit of force [ F ] is kg · m/s 2 , called a Newton (N). Question 3 Part 3 of 3. 10 points. The change in potential energy U of the weight is 1. U = 1 2 ( F m g ) d 2. U = ( F m g ) d 3. U = 2 m g d 4. U = m g d 5. U = m g d 2 correct 6. U = 1 2 (2 F m g ) d 7. U = 2 ( F + m g ) d homework 15 – BAUTISTA, ALDO – Due: Feb 27 2006, 4:00 am 2 8. U = ( F 2 m g ) d 9. U = 3 2 F d 10. U = 2 ( F m g ) d Explanation: If we consider the forces on the pulley we have T m g T As the weight is lifted with no acceleration we have, by force balance, 2 T = m g , or T = F = m g 2 ....
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This note was uploaded on 10/13/2009 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
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