p0415 - 216 CHAPTER 4. WORK AND ENERGY 4.15 Chapter 4,...

Info iconThis preview shows page 1. Sign up to view the full content.

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

Unformatted text preview: 216 CHAPTER 4. WORK AND ENERGY 4.15 Chapter 4, Problem 15 Problem: A block of mass m is dropped from a distance H above a spring-supported surface. The mass of the surface is negligibly small 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 , what is k? Express your answer in terms of m, H , and gravitational acceleration, g . 4 ............... ....................... . ....................... ........................... ................. .. ......................... ......................... ......................... ......................... ......................... ......................... ......................... . .. . .. . . . . . . . . m . . . . . . . . . . . . . . . . . ... ... .. . .. . .. . g = −g k z . . . . . . . . . .. . . .... . ...................... .... .. . . . . . ................................... . ... ............. ... ............................................. ....................................... . . ......................... .. . . . . .. . . .... ..... .. . . ..... . ....... . ....... . ... .. ... . .......... . ....... ...... ...... ........ ....... . . .. ... . .... . ...... ... ...... . . . .......... . .......... ...... ... . . ........ .. ........ .... . . .................................................................................................................................................. .. . . . . . . . . . . . . ..... . . . . . . . .. . . . . . . ... . . . . . ... . . . . ... . . . ............................................................. ............. ...... .... ....... ................................................................................................................................................................................................... .. .................................................................................................................................................................................................. . . .................................................................................................................................................................................................. . . .................................................................................................................................................................................................. ... . . . ... . . ........................................................................................... . . ........................................................................... .................... . .................................................................................................................................................. . . . .. . .. . . . . . . . . . . . ........ ....... . . . . . . . .. .. . ........ ....... H h • • k Solution: 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 unstretched state and the weight is a distance H + h above z = 0. Thus, the initial total energy is Ei = mg (H + h) 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 = mv 2 + kh2 2 2 Conservation of energy tells us that Ef = Ei , wherefore 1 1 mv 2 + kh2 = mg (H + h) 2 2 Solving for v as a function of spring-compression distance, h, yields v (h) = 2g (H + h) − k2 h m Clearly, the speed of the block when it first strikes the surface can be obtained by setting h = 0 in this equation, viz., v (0) = 2gH Now, we are given v (h) = 1 v (0), which tells us that 2 2g (H + h) − Now, if h = 1 H , we have 4 1 kH 2 1 5 gH − = gH 2 16 m 2 Solving for k yields k = 32 =⇒ mg H 2gH = 1 kH 2 16 m k2 1 h= m 2 2gH =⇒ 2g (H + h) − k2 1 h = gH m 2 ...
View Full Document

This note was uploaded on 12/16/2010 for the course AME 301 at USC.

Ask a homework question - tutors are online