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Unformatted text preview: 272 CHAPTER 4. WORK AND ENERGY 4.8 Chapter 4, Problem 8 Problem: A wooden block of mass m rests on a wooden incline, which makes an angle θ to the horizontal. The block is connected by a flexible inextensible cord to a weight of mass 2m as shown. The pulleys are frictionless and have negligible mass. A lower support that held the weight in place has been removed, and the gravitational field is pulling the weight down. The coefficient of kinetic friction between the block and the incline is µk . What is the block’s velocity, v , when it has moved a tangential distance along the incline? Express your answer in terms of , µk , θ and gravitational acceleration, g . z g = −g k ........ ........ .. . .. . . . . . . . .. ... ... ........ .... • • ..... ............ ............ ........ . .. ... .......................................................................................................................... .................................................................................................................................. ...................................................................................................................................................................... . ........ .................. ................................................................................................................................................. ........................................................................................................................................... .. .. 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............................................................................................................................................................................................................................................. .................................................................................................................................................................................................................................................... .. ... ......... .. ...... m • • • 2m θ x So ut on: The f rs s ep n so v ng h s prob em o s es ab sh he re a on be ween he we gh and he b ock Because he eng h of he cab e s cons an he fac ha s he ower pu ey means ha he b ock w rave up he nc ne w ce he d s ance ha ver ca y Deno ng he we gh ’s ve oc y by vw we conc ude ha he b ock’s ve oc y v v = 2vw ve oc es of he wrapped around he we gh drops s w ce vw e I s mos conven en o work n a coord na e sys em a gned w h he nc ne Deno ng he un vec ors norma o and angen a o he surface by n and t respec ve y hey are re a ed o he un vec ors and k n he x and z d rec ons by n = − s n θ + cos θ k and t = cos θ + s n θ k Turn ng o he Pr nc p e of Work and Energy we have U1−2 = T2 − T1 The mo on s ar s from res so ha he n a k ne c energy s T1 = 0 When he b ock has reached a ve oc y v he k ne c energy of he b ock and he we gh s T2 = 1 1 1 1 2 mv 2 + (2m)vw = mv 2 + (2m) 2 2 2 2 1 v 2 2 = 3 mv 2 4 The work done by he forces ac ng on he b ock and he mass s he sum of hree con r bu ons he work done by fr c on on he b ock Uf he work done by grav y on he b ock Ug and he work done by grav y ˜ on he we gh Ug Thus ˜ U1−2 = Uf + Ug + Ug The work done by grav y on he b ock n rave ng a d s ance angen a o he nc ne Ug s Ug = −mg k · t = −mg k · (cos θ + s n θ k) = −mg s n θ 4.8. CHAPTER 4, PROBLEM 8 The work done by friction on the block, Uf , is Uf = −µk N t · t = −µk N 273 ˜ Finally, the work done by gravity on the weight, Ug , is ˜ Ug = −2mg k · The normal force balances the component of the block’s mass that is normal to the incline, which is mg cos θ. Therefore, we have Uf = −µk mg cos θ 1 2 − k = mg Substituting all of these relations into the Principle of Work and Energy yields mg − mg sin θ − µk mg cos θ = Solving for v , we have v= 4 (1 − sin θ − µk cos θ) g 3 3 mv 2 4 =⇒ g (1 − sin θ − µk cos θ) = 32 v 4 ...
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