MIDTERM_02_REVIEW_02_solution

# MIDTERM_02_REVIEW_02_solution - Version 001 MIDTERM 02...

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Version 001 – MIDTERM 02 REVIEW II – turner – (56910) 1 This print-out should have 15 questions. Multiple-choice questions may continue on the next column or page – find all choices before answering. SlidingBlockMITestQ 001(part1of2)10.0points 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. μ M g tan θd 2. μ M g cos θd 3. μ M g sin θd 4. μ M g cos θd correct 5. μ M g tan θd 6. μ M g sin θd Explanation: From the free-body diagram of the block, the normal reaction force N = M g cos θ. The frictional force F f = μ M g cos θ. Since the block moves in a direction opposite to that of friction, W f = F f d = μ M g cos θd. 002(part2of2)10.0points What is the speed of the block at this point? 1. radicalbig 2 g d (sin θ μ tan θ ) 2. radicalbig 2 g d μ (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 10.0points 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

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Version 001 – MIDTERM 02 REVIEW II – turner – (56910) 2 8. 1/2 Explanation: Since A has a circular orbit, m v 2 A R = G m M R 2 , which implies KE A = 1 2 m v 2 A = G m M 2 R . Similarly, for B, m v 2 B 2 R = G M m 4 R 2 , which implies KE B = 1 2 m v 2 B = G M m 4 R . Therefore, KE A KE B = 2 . FanCartMI06x046 004(part1of2)10.0points A fan cart of mass 0 . 84 kg initially has a velocity of vectorv i = ( 0 . 86 , 0 , 0 ) m / s . Then the fan is turned on, and the air exerts a constant force of vector F = (− 0 . 44 , 0 , 0 ) 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: 0 . 66 kg · m / s. Explanation: The change in momentum is given by Δ vectorp = vector F net Δ t = ( (− 0 . 44 , 0 , 0 ) N)(1 . 5 s) = (− 0 . 66 , 0 , 0 ) kg · m / s . So the x component is 0 . 66 kg · m / s . 005(part2of2)10.0points What is the change in kinetic energy of the fan cart over this 1 . 5 s time interval? Correct answer: 0 . 308314 J. Explanation: To find the change in kinetic energy, we need to know the initial and final velocities.
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