Dynamics_Part27

Dynamics_Part27 - Problem Set 4: Problem 1. Problem: As a...

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Unformatted text preview: Problem Set 4: Problem 1. Problem: As a box of mass m approaches a chute, it is moving horizontally with speed V . After it reaches the bottom of a chute, whose length is 2 and whose coefficient of sliding friction is 1 = 1 , 2 the box slides horizontally for a distance where it encounters a spring. The chute is inclined to the horizontal at an angle , and the coefficient of sliding friction of the horizontal surface is 2 = 1 . 3 Determine the value of the spring constant, k, needed to bring the box to a stop at a distance = 1 2 after impacting the spring. Express your answer in terms of m, g, , and V 2 /(g ). . . . . ................... . ................... . ................... ... . . ................... . .. ................................................ . ................................................ . .. ................... . . . . . ............................................. ....................................... . .. .... . ... .. . . ........ ................ ... . . .. . . ......... .................... . .... . . ...................................... ..................................................... . . ......... .. .. . . ...... .................................... .. . ...................................... ..... . . . . ............. ................ . .. ..... .... . ... . .................. .. . 1 . . .. ......... . ............ . ............... .... ............ . . .......................... ....... 1 ... .... .. .................... .... .. .. . .............. 2 ... . .. ..... .. ...... ............................. ....... .. . .. . .. ... .. . .... .. .................. .... . .. . . .. ...... .......................... ....... .. . ............. . .... .......................... .... .. ............... ..... . . ... .......... ...... .. .. ..... .... ...................... .... ................ .... ..... . . .... ....................... ..... .. . .. .... ................... .. .... .......................... ........ .... . ..................... ..... ................................ 1 .. . . .. ................................ ...... .......................... .... ....... ...................... .... 2 ...................... .... ...... 3 ....................... .... ..... ........................ ... ........................ .... ..... .... ........ ............ .... ..... ..... . . ..................... .... .... . .... . ...................... ... .... . . . .. .. .... . . . .. . . . . . . ................. ....................... ..... .... . . . . . . . .. . . . . .. . . . . . ............... . .... ...... ....................... ....... ....... . . .. . . . .. .. . ........ . . . . . . . . ... . . . ....................... .... . . . . .. . . . ........... .. . ....... . .... .... .... .. . . . . . . . ...................... .. . . . . . ............... ...... . . ... . .... .. . .. . . . ..................... ........ ....... .. .... ....... ...................... ............. . .. . . . . .... .. ...... . . . .... ............................................... . .... ..................... ....... . ....................... . . . . .... .............................. .............. . . .... ...................... .... . ................... . .... . .. . .................... . .... ............... . . ................................................ ............................................... ...................... .. ........ ...................... . .......... . . . ..... . ......................... . . . . ............ . .. . .. ........................... ... ................... .. . . ........... . ............ .. .......... . .. ............ ........................ .......... .. ... .. .... ................ ....................... . . ............. ............ ...... ..... ............ ............ .. ......................... . ............ ... .... ... . . . ............ . .. .. ................................................... ... ......................... .... ............ ... ............ .. ......... .... ......................... ............................................................................... .... .............................................................. .... ........... ............ .... . . .............................................................................................. ............ . .. . ...................................................................................................... .. . . . ...... .. . .. . .. . .... . .. . ..... . . ............ . ......... ................................................................................ . .................................................................. ... ............................................... .................................. ................ .. .... m V g = = 2 k Solution: To solve, we first use the Principle of Work and Energy for the box's motion along the chute. This will establish its speed when it reaches the bottom of the chute. Then, we use the Principle of Work and Energy again on the horizontal surface. The forces acting on the box when it's on the chute are as shown below. Clearly, since the box is not accelerating in the direction normal to the chute, N = mg cos . Thus, the magnitude of the friction force is 1 mg cos , and it opposes the motion. The tangential component of the box's weight accelerates it as it moves down the chute, and its magnitude is mg sin . 1 N .......... ... ... ...... .... .. .. ............. ............. ................... ..................... ................................. ............ . ... . .................... ............................... . .. . .... ................ ................................ .. ............... ..................................... . .............. ................................. ............. ............................ ... .. . . ... . . ... . .. .. . . ....................................... . ....................................... ............................. .. ..... .. . .... ................... .. .......................... .................. . ................... . . .. ........ .............. . ... .............. ...................... .. ..... ... . . . ................... . .... ......... . ........ . . . ................... . .. .. . .................... . . ............... . ................ ...... . . . ............ . . .. ................ . ...... .. ............. . ................ . . .. .............. ............... . . .............. . ......... ........... . . .. ............. . .. ............ ............... . .... . . . ... ................. . ............. ... .. . ........... .. . .. . . ............... . .. . ............... .. ................ . . .. . ............. . ................. ... .... ............... ........... . .. ... ............... ............... ..... ..... ................. . ............. ............ .............. ............ .............. . .............. .. . ................... .......... .. ............. ..................... . ...... . . ................... . . . ............. ............ . ...................... . ............... . ............. . .... . .............. . ............... . ... ........... . . . .................................................... ................................................. ............. .. . ..... N mg Since both forces are constant, the work they do as the box traverses the chute is U1-2 = (mg sin - 1 mg cos )(2 ) = U1-2 = 2mg sin - 1 cos 2 where we have used the fact that 1 = 1 . The Principle of Work and Energy tells us that 2 T1 + U1-2 = T2 Here, we have T1 = 1 mV 2 2 ...
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This note was uploaded on 05/12/2010 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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