SolutionsSet05

SolutionsSet05 - Homework Solutions Set 05 Physics 121...

Info iconThis preview shows pages 1–4. Sign up to view the full content.

View Full Document Right Arrow Icon
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 &
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
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
Background image of page 2
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 ( ) " # & & $ %
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 4
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

Page1 / 14

SolutionsSet05 - Homework Solutions Set 05 Physics 121...

This preview shows document pages 1 - 4. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online