problem07_57

University Physics with Modern Physics with Mastering Physics (11th Edition)

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7.57: The two design conditions are expressed algebraically as N 10 66 . 3 4 × = + = mg f ky (the condition that the elevator remains at rest when the spring is compressed a distance y ; y will be taken as positive) and 2 2 1 2 2 1 kx fy mgy mv = - + (the condition that the change in energy is the work fy W - = other ). Eliminating y in favor of k by k y N 10 66 . 3 4 × = leads to k k ) N 10 66 . 3 )( N 10 70 . 1 ( ) N 10 66 . 3 ( 2 1 4 4 2 4 × × + × . N) 10 N)(3.66 10 (1.96 J 10 5 . 62 4 4 4 k × × + × = This is actually not hard to solve for m N 919 = k , and the corresponding x is 39.8 m. This is a very weak spring constant, and would require a space below the operating range
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Unformatted text preview: of the elevator about four floors deep, which is not reasonable. b) At the lowest point, the spring exerts an upward force of magnitude mg f + . Just before the elevator stops, however, the friction force is also directed upward, so the net force is f mg f mg f 2 ) ( =-+ + , and the upward acceleration is 2 2 s m . 17 = m f ....
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  • Force, upward acceleration, upward force, lowest point, design conditions

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