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Unformatted text preview: Version 001 – MIDTERM 02 REVIEW II – turner – (56910) 1 This printout should have 15 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. SlidingBlockMITestQ 001 (part 1 of 2) 10.0 points Suppose a block of mass M is sliding down an inclined plane which makes an angle θ with the horizontal. The coefficient of kinetic friction between the plane and the block is μ (assume tan θ > μ ). The block is initially at rest. What is the work done by friction as the block travels a distance d along the plane? 1. μM g tan θd 2. μM g cos θd 3. μM g sin θd 4. − μM g cos θd correct 5. − μM g tan θd 6. − μM g sin θd Explanation: From the freebody diagram of the block, the normal reaction force N = M g cos θ. The frictional force F f = μM g cos θ. Since the block moves in a direction opposite to that of friction, W f = − F f d = − μM g cos θd. 002 (part 2 of 2) 10.0 points What is the speed of the block at this point? 1. radicalbig 2 g d (sin θ − μ tan θ ) 2. radicalbig 2 g dμ (sin θ − cos θ ) 3. radicalbig 2 g d (sin θ + μ cos θ ) 4. radicalbig 2 g d (tan θ − μ cos θ ) 5. radicalbig 2 g d (sin θ − μ cos θ ) correct 6. μ radicalbig 2 g d (sin θ − μ cos θ ) Explanation: E i = M g d sin θ. The final energy E f = 1 2 M v 2 f . Therefore, energy conservation E i + W f = E f implies, M g d sin θ − μM g cos θ d = 1 2 M v 2 f , or v 2 f = 2 g d (sin θ − μ cos θ ) . Therefore, v f = radicalbig 2 g d (sin θ − μ cos θ ) . PlanetsMITestQ 003 10.0 points Planet A and planet B of equal mass m orbit the same star of mass M in circular trajecto ries of radii r A = R and r B = 2 R respectively. Calculate the ratio of the kinetic energy of A to the kinetic energy of B. 1. 16 2. 1/4 3. 1/8 4. 2 correct 5. 1/16 6. 8 7. 4 Version 001 – MIDTERM 02 REVIEW II – turner – (56910) 2 8. 1/2 Explanation: Since A has a circular orbit, mv 2 A R = GmM R 2 , which implies KE A = 1 2 mv 2 A = GmM 2 R . Similarly, for B, mv 2 B 2 R = GM m 4 R 2 , which implies KE B = 1 2 mv 2 B = GM m 4 R . Therefore, KE A KE B = 2 . FanCartMI06x046 004 (part 1 of 2) 10.0 points A fan cart of mass 0 . 84 kg initially has a velocity of vectorv i = ( . 86 , , ) m / s . Then the fan is turned on, and the air exerts a constant force of vector F = (− . 44 , , ) N on the cart for 1 . 5 s. What is the change in the x component of momentum of the fan cart over this 1 . 5 s time interval? (Since the force is only applied along the x direction, the other components of momentum will not change.) Correct answer: − . 66 kg · m / s. Explanation: The change in momentum is given by Δ vectorp = vector F net Δ t = ( (− . 44 , , ) N)(1 . 5 s) = (− . 66 , , ) kg · m / s ....
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This note was uploaded on 03/20/2012 for the course PHY 303K taught by Professor Turner during the Spring '08 term at University of Texas.
 Spring '08
 Turner
 Physics

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