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Unformatted text preview: homework 11 – FIERRO, JEFFREY – Due: Feb 25 2008, 11:00 pm 1 Question 1, chap 7, sect 4. part 1 of 1 10 points Mary weighs 517 N and she walks down a flight of stairs to a level 4 . 7 m below her starting point. What is the change in Mary’s potential en ergy? Correct answer: 2429 . 9 J (tolerance ± 1 %). Explanation: We can let the initial potential energy be 0. The change in potential energy is Δ U = U f U i = mg ( h ) = (517 N)(4 . 7 m) = 2429 . 9 J . Question 2, chap 7, sect 4. part 1 of 2 10 points Given: G = 6 . 67259 × 10 − 11 N · m 2 / kg 2 M E = 5 . 98 × 10 24 kg R E = 6 . 37 × 10 6 m A satellite has a mass of 84 kg and is located at 1 . 49 × 10 6 m above Earth’s surface. What is the potential energy of the satellite Earth system? Correct answer: 4 . 26435 × 10 9 J (tolerance ± 1 %). Explanation: Given : m = 84 kg , M E = 5 . 98 × 10 24 kg , R E = 6 . 37 × 10 6 m , and h = 1 . 49 × 10 6 m . The radius of the satellite’s orbit is r = R E + h. The potential energy is given by E g = GmM E R E + h = 6 . 67259 × 10 − 11 N · m 2 / kg 2 · (84 kg) (5 . 98 × 10 24 kg) 6 . 37 × 10 6 m + 1 . 49 × 10 6 m = 4 . 26435 × 10 9 J . Question 3, chap 7, sect 4. part 2 of 2 10 points What is the magnitude of the gravitational force exerted by Earth on the satellite? Correct answer: 542 . 538 N (tolerance ± 1 %). Explanation: The force acting on the satellite is F = GmM E ( R E + h ) 2 = E g R E + h = 4 . 26435 × 10 9 J 6 . 37 × 10 6 m + 1 . 49 × 10 6 m = 542 . 538 N . Question 4, chap 7, sect 4. part 1 of 1 10 points An object is dropped from rest from a height 5 . 3 × 10 6 m above the surface of the earth. The acceleration of gravity is 9 . 81 m / s 2 . If there is no air resistance, what is its speed when it strikes the earth? Correct answer: 7 . 53393 km / s (tolerance ± 1 %). Explanation: Let : h = 5 . 3 × 10 6 m . g = GM E R 2 E and the potential energy at a distance r from the surface of the earth is U ( r ) = GM E m r . homework 11 – FIERRO, JEFFREY – Due: Feb 25 2008, 11:00 pm 2 Using conservation of energy to relate the initial potential energy of the system to its energy as the object is about to strike the earth ( K i = 0), K f + U f U i = 0 K ( R E ) + U ( R E ) U ( R E + h ) = 0 1 2 mv 2 GM E m R E + GM E m R E + h = 0 . Solve for v : v = radicalBigg 2 parenleftbigg GM E R E GM E R E + h parenrightbigg = radicalBigg 2 g R E parenleftbigg h R E + h parenrightbigg = radicalBigg 2 (9 . 81 m / s 2 ) (6 . 37 × 10 6 m) 5 . 3 × 10 6 m + 6 . 37 × 10 6 m × radicalbig 5 . 3 × 10 6 m · 1 km 1000 m = 7 . 53393 km / s ....
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 Spring '08
 Turner
 Energy, Friction, Kinetic Energy, Potential Energy, Work, Light, Correct Answer, JEFFREY

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