Work and Energy-solutions

# Work and Energy-solutions - le (txl65) Work and Energy...

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le (txl65) – Work and Energy – mohanty – (81440) 1 This print-out should have 26 questions. Multiple-choice questions may continue on the next column or page – fnd all choices beFore answering. 001 10.0 points A 12 g bullet is accelerated in a ri±e barrel 72 . 1 cm long to a speed oF 780 m / s. Use the work-energy theorem to fnd the average Force exerted on the bullet while it is being accelerated. Correct answer: 5062 . 97 N. Explanation: Let : m = 12 g = 0 . 012 kg , x = 72 . 1 cm , and v = 780 m / s . The kinetic energy oF the bullet beFore it is fred is 0 J since it is not moving. At the moment that it leaves the ri±e barrel, its kinetic energy is K = 1 2 mv 2 = 1 2 (0 . 012 kg) (780 m / s) 2 = 3650 . 4 J . Since the work done on the bullet is equal to its change in kinetic energy, this is also the work that is done on the bullet: W = 1 2 mv 2 = 3650 . 4 J . Also, the work is given by W = F avg x F avg = W x = 3650 . 4 J 0 . 721 m = 5062 . 97 N . 002 (part 1 of 2) 10.0 points A bead slides without Friction around a loop- the-loop. The bead is released From a height h From the bottom oF the loop-the-loop which has a radius r . h r Which oF the Following diagrams best rep- resents the kinetic energy oF the bead versus time? 1. t K 2. t K correct 3. t K 4. t K 5. t K 6. t K Explanation: The bead reaches a maximum kinetic en- ergy each time it reaches the bottom oF the loop. Consequently, the only curve which fts this criteria is one which has two equally high maxima. Also, the kinetic energy must be zero at t = 0.

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le (txl65) – Work and Energy – mohanty – (81440) 2 003 (part 2 of 2) 10.0 points Which of the following could represent the gravitational potential energy of the bead ver- sus time? 1. t U 2. t U 3. t U 4. t U 5. t U 6. t U correct Explanation: Since the kinetic energy plus potential en- ergy is constant, this curve must be the inverse of the curve in Part 1. 004 10.0 points An ore car of mass 35000 kg starts from rest and rolls downhill on tracks from a mine. At the end of the tracks, 20 m lower vertically, is a horizontally situated spring with constant 2 . 1 × 10 5 N / m. The acceleration of gravity is 9 . 8 m / s 2 . Ignore friction. How much is the spring compressed in stop- ping the ore car? Correct answer: 8 . 0829 m. Explanation: Energy is conserved, so the change of po- tential energy from when the car is at rest to when it just hits the spring is mg h = 1 2 mv 2 . The kinetic energy is then converted to po- tential energy in the spring as the cart comes to rest. When the spring is fully compressed by an amount d , all of the kinetic energy has been converted to potential energy so 1 2 mv 2 = 1 2 k d 2 . Thus, 1 2 k d 2 = mg h, and solving for d we have d = r 2 mg h k = R 2 (35000 kg) (9 . 8 m / s 2 ) (20 m) (2 . 1 × 10 5 N / m) = 8 . 0829 m .
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## This note was uploaded on 04/05/2011 for the course SD ds taught by Professor Sdsd during the Spring '11 term at Tarrant County.

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Work and Energy-solutions - le (txl65) Work and Energy...

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