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Unformatted text preview: 4.18. CHAPTER 4, PROBLEM 18 289 4.18 Chapter 4, Problem 18 Problem: A block of mass m is dropped from a distance H above a spring-supported surface. The mass of the surface is ms = m and the spring constant is k. If the block’s speed, v , is half of its value when it first strikes the surface when the weight has compressed the spring through a distance h = 1 H , 4 what is k ? Express your answer in terms of m, H , and gravitational acceleration, g . ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ......................... ........................ . .. . . .. . . . . . . . . m . . . . . . . . . . . . . . . . . . ... .. . .. . . g = −g k z . . . . . . . . . .. . .................. . ...................................... . . .... ..... . . . ..................................... . ....... . ..... ..... . ..... . . . . ......................................... ............................. .. ..... . . . ........... .......... ..... ......... .. ........ ..... . . ........ ....... .. . ..... .. ........... ........... . . . ....... ........... . ........... . . ... . ........ ........... . .............................................................................................................................................................................................................................................................................................. .. ..................................................................................................... . .................................................................................................................................................... ..... ...................................................................................................................... .................................................................................................................................................. . .................................................................................................................................................................................................. . ................................................................................................................................................... . .................................................................... ......... . .. .. ............ . .. . . . . . . . . . . . . . . . . . . . . . . .. ...................................................................................................... . . . . .. . . .. . . . . . . . . . . . . ........ ....... . . . . . . . .. .. . ........ ....... H h • • k Solution: The first thing we must do is determine the equilibrium displacement of the spring. That is, because of the weight of the surface, the spring is compressed below its unstretched position. Calling this position zeq , obviously ms g ms g = kzeq =⇒ zeq = k Because only conservative forces are acting, i.e., gravity and the spring, total energy is conserved. In the initial state, there is no motion, the spring is in its equilibrium state, the weight is a distance H + h above z = 0 and the surface is a distance h above z = 0. Thus, the initial total energy is 12 Ei = mg (H + h) + ms gh + kzeq 2 When the weight has compressed the spring though a distance h, its speed is v . The total energy at this time is 1 1 Ef = (m + ms ) v 2 + k (h + zeq )2 2 2 Conservation of energy tells us that Ef = Ei , wherefore 1 12 1 2 (m + ms ) v 2 + k (h + zeq ) = mg (H + h) + ms gh + kzeq 2 2 2 Regrouping terms, we find 1 (m + ms ) v 2 2 1 2 = mgH + (m + ms ) gh + k zeq − (h + zeq )2 2 1 2 = mgH + (m + ms ) gh + k zeq − (h + zeq )2 2 1 = mgH + (m + ms ) gh + k (zeq + h + zeq ) (zeq − h − zeq ) 2 1 = mgH + (m + ms ) gh − kh (h + 2zeq ) 2 1 = mg (H + h) − kh2 2 Now, using the fact that zeq = ms g/k , there follows 1 1 ms g (m + ms ) v 2 = mgH + (m + ms ) gh − kh h + 2 2 2 k 290 CHAPTER 4. WORK AND ENERGY We can determine the speed of the block at the moment it strikes the plate by setting ms = 0 (the plate surface has not yet begun to move) and h = 0 (the spring has not been compressed beyond its equilibrium position) from this equation. Thus, calling this velocity vo , 1 mv 2 = mgH 2o =⇒ vo = 2gH Finally, we are given ms = m, v = 1 vo and h = 1 H . Combining these facts, we conclude that 2 4 1 (2m) 2 1 2 2 2gH = 5 1 mgH − k 4 2 1 h 4 2 =⇒ mg H 1 5 1 mgH = mgH − kh2 2 4 8 Finally, solving for k gives k = 24 ...
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This note was uploaded on 05/11/2011 for the course AME 301 taught by Professor Shiflett during the Spring '06 term at USC.

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