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SolutionsSet05 - Homework Solutions Set 05 Physics 121...

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Homework Solutions Physics 121 Set 05 Spring 2008 - 1 - WeBWorK Problem 1 The force F is given by F ( x ) = 6.4 x 2 + 4.3 x . In order to determine the potential energy at x = 3.0 m we need to calculate the path integral of F along the path that connects x = 0 m with x = 3.0 m. In the region between x = 0 m with x = 3.0 m the force is positive and force and displacement are pointing in the same direction. The work done by the force is thus positive. Consider a small path segment at x : dx . The work done by F when we move from x to x + dx is equal to dW = (6.4 x 2 + 4.3 x ) dx . The potential energy at x = 3.0 m is thus equal to U 3 ( ) = ! 6.4 x 2 + 4.3 x ( ) dx = ! 6.4 3 x 3 + 4.3 2 x 2 " # \$ % 0 3 = ! 77 J x = 0 x = 3 &

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Homework Solutions Physics 121 Set 05 Spring 2008 - 2 - WeBWorK Problem 2 The initial mechanical energy of an object of mass m fired from the surface of the earth with a velocity v i is equal to E i = 1 2 mv i 2 ! G mM E R E At an infinite distance, the potential energy will be zero, and the mechanical energy will thus be equal to the kinetic energy of the object: E f = 1 2 mv f 2 Since mechanical energy is conserved, we require that 1 2 mv i 2 ! G mM E R E = 1 2 mv f 2 or v i = f 2 + 2 G M E R E
Homework Solutions Physics 121 Set 05 Spring 2008 - 3 - WeBWorK Problem 3 We will take the top of the spring (in its rest position) to be the position where the potential energy is equal to 0 J. When the elevator is at rest a distance d above the spring, it will only have potential energy ( E = mgd ). When the cable snaps, the elevator will move down, but its downwards motion is slowed down by the clamps on the guide rails that provide a constant frictional force f . The work done by the friction force up to the point that the elevator reaches the spring is – fd . The mechanical energy of the system at this point will thus be equal to mgd fd . At this position, all the mechanical energy is in the form of kinetic energy of the elevator. Consider what happens when the spring is compressed by a distance x . Assuming that this is the maximum compression (that is, the elevator has come to rest), we know that the total mechanical energy of the system is – mgx + (1/2) kx 2 . During the compression, the work done by the friction force is – fx , and the total mechanical energy must therefore also be equal to mgd fd fx . We thus require that ! mgx + 1 2 kx 2 = mgd ! fd ! fx This equation can be rewritten as 1 2 kx 2 + f ! mg ( ) x + f ! mg ( ) d = 0 which can be solved for x : x = ! f ! mg ( ) ± f ! mg ( ) 2 ! 4 f ! mg ( ) d 1 2 k " # \$ % k = mg ! f k 1 ± 1 + 2 kd mg ! f ( ) " # & & \$ %

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SolutionsSet05 - Homework Solutions Set 05 Physics 121...

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