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Unformatted text preview: This printout should have 20 questions. Multiplechoice questions may continue on the next column or page – find all choices before answering. 001 10.0 points A 6 . 8 kg object hangs at one end of a rope that is attached to a support on a railroad boxcar. When the car accelerates to the right, the rope makes an angle of 42 ◦ with the vertical The acceleration of gravity is 9 . 8 m / s 2 . a 6 . 8 kg 4 2 ◦ Find the acceleration of the car. ( Hint: vectora object = vectora car ) 1. 8.82396 2. 3.37441 3. 10.1482 4. 3.95946 5. 4.77978 6. 6.36419 7. 7.6566 8. 6.86203 9. 9.46375 10. 8.51901 Correct answer: 8 . 82396 m / s 2 . Explanation: Given : m = 6 . 8 kg , θ = 42 ◦ , and g = 9 . 8 m / s 2 . T T sin θ T cos θ θ mg Vertically summationdisplay F y = T cos θ − mg = 0 T cos θ = mg . (1) Horizontally, summationdisplay F x = T sin θ = ma. (2) Dividing Eqs 1 and 2, we have T sin θ T cos θ = a g tan θ = a g a = g tan θ = ( 9 . 8 m / s 2 ) tan42 ◦ = 8 . 82396 m / s 2 . 002 10.0 points A 30 kg child on a 2 m long swing is released from rest when the swing supports make an angle of 21 ◦ with the vertical. The acceleration of gravity is 9 . 8 m / s 2 . If the speed of the child at the lowest po sition is 1 . 44323 m / s, what is the mechani cal energy dissipated by the various resistive forces ( e.g. friction, etc.)? 1. 7.81095 2. 9.48618 3. 11.3474 4. 10.0617 5. 17.592 6. 15.0607 7. 7.72821 8. 18.4309 9. 20.7165 10. 18.6802 Correct answer: 7 . 81095 J. Explanation: The total energy at the top of the swing is potential, and the total energy at the bottom is kinetic. At the top of the swing, the swing is a ℓ cos θ below the support, so it is ℓ (1 − cos θ ) above its lowest point. The mechanical energy lost due to friction would thus be Δ E = E top − E bottom midterm 02 – Turner – (58185) 1 Version 132/ACABA – midterm 02 – Turner – (58185) 2 = mgh − 1 2 mv 2 = (30 kg)(9 . 8 m / s 2 )(0 . 132839 m) − 1 2 (30 kg)(1 . 44323 m / s) 2 = 7 . 81095 J 003 10.0 points A cart loaded with bricks has a total mass of 22 . 6 kg and is pulled at constant speed by a rope. The rope is inclined at 26 ◦ above the horizontal and the cart moves 11 . 6 m on a horizontal floor. The coefficient of kinetic friction between ground and cart is 0 . 5 . The acceleration of gravity is 9 . 8 m / s 2 . How much work is done on the cart by the rope? 1. 0.481461 2. 2.30049 3. 0.731498 4. 1.03273 5. 0.962998 6. 1.14434 7. 0.781245 8. 2.39453 9. 1.75933 10. 0.907327 Correct answer: 1 . 03273 kJ. Explanation: F 2 6 ◦ 22 . 6 kg μ = 0 . 5 Summing the components of the forces along the vertical direction provides summationdisplay F y = F sin θ + N − mg = 0 ....
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 Spring '08
 Turner
 Force, Mass, kg, Standard gravity

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