rachelg1456
  • rachelg1456
Bob has a tree house at a height of 20 ft. When he's feeling adventurous he uses a rope to climb the tree house instead of the ladder. If the weight of the rope is 32 lb, then how much work must Bob do to pull the rope up once he has entered the house? A) 173.00 ft-lb B) 235.29 ft-lb C) 320.00 ft-lb D) 383.54 ft-lb Please help me by working through the problem with me! I suck at math and physics :(
Physics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
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shamim
  • shamim
U know the centre of mass of the rope is just middle of the rope
shamim
  • shamim
Can u tell me the height of the tree house frm centre of mass of the rope
shamim
  • shamim
Work done=mgh=?

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anonymous
  • anonymous
The Correct Choice: C Reason: I'm not sure if you know integration in math or not. If you Don't know: Bob needs to pull the center of mass of the rope up. Where is the center of mass of the rope? For any symmetric object it is located at its center. A straight rope can be taken as a symmetric object. So it's center (at 10 ft from the tree house) is also its Center of Mass. W=(mg).h Work needed = (Weight of the rope).(The distance the center of mass must be moved vertically) \[W=32lb \times 10ft = 320 ft-lb\] If you know: To pull a very thin horizontal slice of the rope(dy) which is at a distance from the tree house (y), the work (W) Bob need to do is: \[\frac{ mg }{ h }.ydy\] Now if you solve the definite integral between 0 and 20 ft, you'll have the answer: W=\[\int\limits_{0}^{20}\frac{ mg }{ h }.ydy= \frac{ mg }{ h }\int\limits_{0}^{20}ydy= \frac{ mg }{ h }.\frac{y ^{2} }{ 2 }= \frac{ 32lb }{ 20ft }\times \frac{20ft ^{2} }{ 2 }=320ft-lb\] Do Great,

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