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Unformatted text preview: 4.4. CHAPTER 4, PROBLEM 4 257 4.4 Chapter 4, Problem 4 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 kinetic friction is µ1 = 1 , 2 the box slides horizontally for a distance where it encounters a spring. The chute is inclined to the horizontal at an angle θ, and the coefficient of kinetic friction on the horizontal surface is µ2 = 1 . 3 Determine the value of the spring constant, k, needed to bring the box to a stop at a distance δ = 1 2 after impacting the spring. Express your answer in terms of m, , θ, gravitational acceleration, g = |g|, and the dimensionless quantity V 2 /(g ). . . . .................. ..... . ....................... . .. .. .. .. .. . ............................................... . .. . .............. .. .. . ................... . ... ................................................ . . . .................................................. .............................. ... .. ......................... . . .... . ... ... .. .. .. . . ............................................. ........ . . . . .......................................... . .. .. . . ......................................................................... ....................................................... . .... . . . . . . . .. . ...................................... . . ............................................... . ............. ..... . ...... . . . .. .......................... 1 . . . .... . .. ... . . ...................... . ............ ........................ . .. ................................. ............. 1 ......................... ............................ ... .. . ..... ..... 2 .. ....... ............... ... .. ....... . .. ... . . ........................... . .. ............................ ...................... . .. . .... .. ...... .......................... .. . .. ..... . ........ .. ..... .......................... .... .... ..... .......................... . . ... .. .......................... .... . .... ......................... ...... ... .... ............................ .. .... .... ........................... .. ...... .. .... .. .......................... .... ....... ... . . ... .... .......................... .. ................. . ................... 1 .... .. .... ............................. .................. . .. . ..... .... ........................... .... .......................... 2 ... . . . ............................. 3 .... .... .......................... . ..... . ........ ........................... .... .... .......................... ...... . .... . .. . .. ......... . . ... ..... .......... . ............................ .. ......................... . .... . .... .. . .. . ........................... ... . .. .. ........... . .... ... .. ............ . ........................... . . .......................... .... . . . ............ . ........ .. .. ..... . .. ............ . .. .... .. .. .. ............ .. . . .. ....... .. . . ............. ............................ .. . .... . .. ... ......................... . . ..... ......... .. .. .... . . . . . . .............. .... .... . .. .. . .... .. .. .. ............ . .. .. ........................... .... . . . ............ ....................... .................... .... . ..... ....... .. .... ... . . . ............ ............................ . .... . .. . .............. .... .. ..................... .. . . ... . .... .. .... ............................ ... . .. .. ............ . . ......................... . .. . . .... .................... . .... ............................................................................ ............ .. .. ............................................................................................................................... ..... .............................................................................................................. .... . . .... ................... .... .... ... ..... ... .. . . ... . . . .. .... .... .... ............... .......... .................................................................. .... ........................................................................................................................................ .... ................................................................................................................................... . ...... ............................................................... .. .... .............................................................................................. .... ...................................................................... .. . . . . ............................................... .... .... ..... ... . .... .. . . ... . ........ .... .......... ... . .......... ....... . . . .. .. . .. . . . . . . m V θ g µ= µ= 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 θ. Thus, the magnitude of the friction force is µ1 mg cos θ, 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 θ. µ1 N ........... . .. .... .. ..... ... .. .... ............. . ................... ... .. . ....... ... . ....................... ........................................ ..... . ............... . .............................. .................. ....................................... . .... .............. ................................. .............. .............................. . .............. .................................. ... ..... ................ .. . .. ....................................... .... ................................. . .. ... .. . . .............................. ........................... .............. . .. .. . .. ..................... ...... ........ .. .... ................ ... .... .... . .. .................... .... ................... .. . . ... .. . . .. .... . . .......................... . . ..................... . . .................... ..... ........ . .............. . ............... .. . ............ . . .......... ................. . . ................ . .. ............. . .............. . ............. ............. . ..... ...... .. .............. . .. ............. . ............. .............. . ............. .. ... .. . ............. ............. .. ................ .. . .. ..... ..... . ............... .. .. . ................ ............ . ......... . ............ .......... . ...... .. .. . .............. ............. .............. ................ ............. ................. . .... .. ................ .............. ................. . ... ..... ............. ........... . ................. .. . . .............. . ....... .. . .................... . ...................... . ........... . ............ . . .................... . ............ ............ . ............... . .... ...... . ............. . ..... .. . . . .. . .................................................... ..................................................... . .......... . ...... N mg θ Since both forces are constant, the work they do as the box traverses the chute is U1−2 = (mg sin θ − µ1 mg cos θ)(2 ) =⇒ U1−2 = 2mg sin θ − 1 cos θ 2 where we have used the fact that µ1 = 1 . The Principle of Work and Energy tells us that 2 U1−2 = T2 − T1 Here, we have T1 = 1 mV 2 2 258 CHAPTER 4. WORK AND ENERGY Denoting the box’s velocity at the bottom of the chute by v2 , we have 1 mV 2 + 2mg 2 sin θ − 1 cos θ 2 = 1 mv 2 22 On the horizontal surface, the only tangential force acting until the box encounters the spring is friction, which is µ2 mg to the left. The spring force is Fe = −kx and the work done in compressing the spring a distance δ is − 1 k δ 2 . Thus, the work done as the box moves from the beginning of the horizontal surface 2 until it comes to rest is 1 U2−3 = −µ2 mg ( + δ ) − k δ 2 2 The Principle of Work and Energy tells us that U2−3 = T3 − T2 Now, since the box comes to rest, its speed is v3 = 0, so that its kinetic energy is T3 = 0. Thus, 1 1 2 mv2 − µ2 mg ( + δ ) − k δ 2 = 0 2 2 2 Substituting for 1 mv2 from above, this equation becomes 2 1 mV 2 + 2mg 2 We are given δ = 1 2 sin θ − 1 cos θ 2 1 = µ2 mg ( + δ ) + kδ 2 2 and µ2 = 1 , wherefore 3 3 2 1 +k 2 1 2 2 1 1 µ2 mg ( + δ ) + kδ 2 = mg 2 3 Combining these two equations yields 1 1 mg + k 2 8 Solving for k, we conclude that k= 4mg 2 = 1 1 mg + k 2 8 2 = 1 mV 2 + 2mg 2 sin θ − 1 cos θ 2 4 sin θ − 2 cos θ − 1 + V2 g ...
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