p0410 - 208 CHAPTER 4. WORK AND ENERGY 4.10 Chapter 4,...

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Unformatted text preview: 208 CHAPTER 4. WORK AND ENERGY 4.10 Chapter 4, Problem 10 Problem: A block of mass m moves in a vertical slot as shown. A spring of unstretched length L and spring constant k is attached to the block. The coefficient of sliding friction between the block and the surface on the right side of the slot is µs = 1 . The left surface is frictionless. 3 (a) Verify that the magnitude of the normal force exerted by the right slot surface on the block is √ N = kL 1 − L/ L2 + z 2 . (b) How much work is done when the block moves a distance 4 L downward? NOTE: The tangential 3 component of the spring force and the friction force both vary with distance moved, so you will have to do an integral for both. √ (c) If, after moving a distance 4 L, the block’s speed is v = λgL, where λ = 4 n3, what is the 3 3 value of the spring constant, k ? . . . . .. .. .. . ........... .......... .. . .. ............ ............ . . . . . . .... ....... . . ...... . .. .. .. .... ...... ...... .. ..... .... ...... ...... ...... .. .. . .. ...... ...... ... . .. . .. .. ..................... .. .. .. .............................. .... .. . .. .................................. .. ....... . ..... ....... ......... ........ .................................... .... . .. .. .................................... . . . . ........... .. .. . .............. . .. ..... .................. ....................... ................... .. .................... ... .. . . .. .. . ...................... .. . .. . . ....... . . .. . ........ .................................................................................... .... ... ................................. ..... ... ... ................................... . . . .. . . .. ........ .. ................. . .. .................................... ... ... ... ... .................... . .. . . . . .. . ...... .. . . ................. . .. .. .. .... .. .. . .. .. .. . ........ .. . .. ................ .................................... .. ... .. . .. . ............................... ........ ................................... . ........ . . ... ........ .. . .. . . . ..... .. . ......... . . . ........ . . ........ . . . ...... . ...... . .. . .. ....... .. . ...... .. .. ...... . .. . ...... .. .. . ...... ...... . .. . .. .. .. . ..... .... ...... ...... . .. ...... . . .. . .. . . .. . .. . . ..... ...... . ...... . .. ....... .. . ...... ... . .... .. . . ...... . . . .. .. . ...... ...... . .. .. . ...... ...... . . ...... .. .. . .. .. ... ...... .. . .. . ..... . .. ...... ...... .. ... . ...... .. .. . .... ...... . 4 . ...... . .. .. ...... . ... . . ...... .. . . ...... ...... . . . ...... .. .. .. .. .. .. ...... . . ...... .... . ...... ......... .. .. 3. ...... .. .. ...... ...... .. .. ...... . . . ...... .. .. ...... .. . . . . ...... . . .. .. ...... .. . . .. .. ...... . ...... . . ...... .. ........ ...... . . . . .. .... . . . . .. .... . . .. . ...... ...... . . .. . .. .. .. .. . . ...... . . .... ..... . . . ........ . ............................... . . ................................ . .. . . .................................. .. . .. . . .. . .. .. ... .. . . . . .................................... .. . . .. . ..... .................................. ..... ................... .. . ..... .. .. ..................................... . ..... ................................... ..... .... . ................................. . .. . .. . . ............... . . .. .................................... . .................................... . ............................ .. .. . . ... .. . .. .... . .. .................................. ...... . . ....... ... .. .. ... .. . .. . .. .. . ...... .. . ... ...... ....... ...... ..... .. . .... . .... .... . . .. . ..... . L • k • These integrals will be helpful: √ √ z dz z 2 + L2 dz z 2 + L2 = √ z 2 + L2 √ z 2 + L2 x L z • g = gk = n z+ Now, we know that Solution: (a) Because the block is constrained to move vertically, the horizontal forces must balance. Denoting the length of the spring in its stretched state by ξ , the magnitude of the spring force is Fs = −k (ξ − L), and it acts in the direction parallel to the spring as shown in the figure below. Thus, balancing horizontal forces tells us that N = k(ξ − L) cos θ ξ= z 2 + L2 and L cos θ = √ 2 + L2 z Fs µN . .. . . ... .. .... .. ... . .. .. .. . .. . . .. .. .. .. . ... s .... . . .. . . .. . .. . . ...... . ...... . . ........................................... .. . . . ... . . . . . . . . . . . . . .. . .. . .......................................... . . .. . . . . . .. .. . .. ................................................ . . . . . . . . ... . . . ................................................... ... .. . . . .. . .. .... . . . ................................................... . . . ................................................... . . . .......................................... . ......................................... .... . . . .............. ........ . .................................................. . . .............................. . . .................................................. . ....................................................... .. . . .. ............................................. .......... . . ...... ...... .................................................... . . .. . . . . ................................................... . . ........... . . .................................................. . ..... . .................................................. ............................... .................................................. . .......................... . . .................................................. . . . . . ................................................. . .. . .. . .................................................. . ...................................... .............. . ............. .. . .......................... . . . . . . . . .. . ... .. .. .. .. . . θ N mg Therefore, the normal force is N =k L L z 2 + L2 − L √ = kL 1 − √ z 2 + L2 L2 + z 2 4.10. CHAPTER 4, PROBLEM 10 209 (b) There are three forces that do work, viz., the spring force, gravity and friction. Proceeding one by one, we have the following. Spring Force. The geometry tells us that the vertical component of the spring force, T , is z T = −k (ξ − L) sin θ and sin θ = √ L2 + z 2 Thus, we have Lz T = −k z − √ L2 + z 2 So, the work done by the spring force is 4 3L Us =− = −k 0 Lz k z−√ L2 + z 2 4 L 3 2 1 dz = −k z 2 − L L2 + z 2 2 L2 + L2 = −kL2 z= 4 L 3 z =0 1 2 −L 16 +1 9 85 2 − + 1 = − kL2 93 9 Gravity. Since the gravitational force is mg and the displacement is 4 Lk, the work done by gravity is 3 Wg = mg k · 4 3L 4 Lk 3 = 4 mgL 3 z= 4 L 3 z =0 Friction. The friction force, µs N , opposes the motion and the work done friction is Uf =− L µs kL 1 − √ L2 + z 2 4 L−L n 3 4 L+ 3 dz = −µs kL z − L n z + 16 + 1 L2 9 L2 + z 2 0 = −µs kL So, the total work done is U1−2 + L nL = −µs kL2 4 − n3 3 4 − n3 3 2 4 = Us + Ug + Uf = − kL2 + mgL − µs kL2 9 3 2 4 4 = mgL − kL2 + µs − n3 3 9 3 (c) The Principle of Work and Energy tells us that 1 1 2 U1−2 = T2 − T1 = mv 2 − 0 = mλgL = n3 mgL 2 2 3 √ where we make use of the given facts that v = λgL, where λ = 4 n3. Thus, using the result of Part (b), 3 and the given value of µs = 1 , there follows 3 21 4 mgL − kL2 + 3 93 Rearranging terms yields 1 2 (2 − n3)mgL = (2 − n3)kL2 3 3 Therefore, the spring constant, k , is mg k=2 L 4 − n3 3 = 2 n3 mgL 3 ...
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This note was uploaded on 12/16/2010 for the course AME 301 at USC.

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