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Unformatted text preview: reyes (mr29667) – Homework 5 – chelikowsky – (59005) 1 This printout should have 23 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 (part 1 of 2) 10.0 points A block of mass m is accelerated across a rough surface by a force of magnitude F that is exerted at an angle φ with the horizontal, as shown above. The frictional force on the block exerted by the surface has magnitude f . f F φ m What is the magnitude of the acceleration vectora of the block? 1.  vectora  = F cos φ m 2.  vectora  = F − f m 3.  vectora  = F cos φ − f m correct 4.  vectora  = F m 5.  vectora  = F sin φ − mg m Explanation: From Newton’s second law of motion, the acceleration is the total force in the horizontal direction divided by the mass. There are two forces in the horizontal direction: one is the friction force; the other is the horizontal com ponent of the dragging force F , but they are in the opposite directions, so the acceleration of the block is  vectora  = F cos φ − f m . 002 (part 2 of 2) 10.0 points Which of the following expressions for the coefficient of friction is correct? 1. μ = mg − F cos φ f 2. μ = f mg − F sin φ correct 3. μ = f mg − F cos φ 4. μ = f mg 5. μ = mg f Explanation: By definition, the coefficient of kinetic fric tion is the ratio of the friction force and the normal force in the vertical direction. And it is easy to see that the normal force in the vertical direction is just N = mg − F sin φ, So the coefficient of friction is μ = f mg − F sin φ . 003 10.0 points The suspended m 1 mass on the right is moving up, the m 2 mass slides down the ramp, and the suspended m 3 mass on the left is moving down. There is friction between the block and the ramp, with the coefficient of the kinetic friction μ . The acceleration of gravity is g and the acceleration of the three block system is a . The pulleys are massless and frictionless. m 2 μ θ m 3 m 1 reyes (mr29667) – Homework 5 – chelikowsky – (59005) 2 What is the equation of motion of the three block system? 1. m 3 g + m 2 g (cos θ − μ sin θ ) − m 1 g = ( m 1 + m 3 ) a 2. m 3 g + m 2 g (sin θ − μ cos θ ) − m 1 g = ( m 1 + m 3 ) a 3. m 3 g + m 2 g (cos θ − μ cos θ ) − m 1 g = ( m 1 + m 3 ) a 4. m 3 g + m 2 g (sin θ − μ sin θ ) − m 1 g = ( m 1 + m 3 ) a 5. m 3 g + m 2 g (1 − μ ) − m 1 g = ( m 1 + m 3 ) a 6. m 3 g + m 2 g (cos θ − μ cos θ ) − m 1 g = ( m 1 + m 2 + m 3 ) a 7. m 3 g + m 2 g (sin θ − μ cos θ ) − m 1 g = ( m 1 + m 2 + m 3 ) a correct 8. m 3 g + m 2 g (1 − μ ) − m 1 g = ( m 1 + m 2 + m 3 ) a 9. m 3 g + m 2 g (sin θ − μ sin θ ) − m 1 g = ( m 1 + m 2 + m 3 ) a 10. m 3 g + m 2 g (cos θ − μ sin θ ) − m 1 g = ( m 1 + m 2 + m 3 ) a Explanation: Basic Concept: F net = ma negationslash = 0 Solution: The acceleration a of each mass is the same, but the tensions in the two strings will be different.will be different....
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This note was uploaded on 07/24/2010 for the course PHY 59005 taught by Professor Chelikowsky during the Fall '09 term at University of Texas.
 Fall '09
 Chelikowsky

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