kingwinner Posted May 3, 2006 Report Share Posted May 3, 2006 I have some questions about Hooke's law. I have posted my thoughts, but getting stuck at the end. Can someone please help me? I would appreciate! :shrug: 1) An archer pulls her bow string back 0.400m by exerting a force that increases uniformly from 0 to 230N. How much work is done in pulling the bow? F=kx, so the equivalent spring constant is 575N/m. The elastic potential energy is (1/2)kx^2=(1/2)(575)(0.400)^2=46J. So is the work done by the archer 46J? I am not sure about that because I was told that WORK DONE and ENERGY are not entirely the same thing..... 2) Total spring constant=65.0x4=260N/mAmount of stretch from equilibrium position=0.365mF=kx=(260)(0.365)=94.9NNow do I have to divide this by 2 to get 47.4N? I think this is getting really TRICKY. Is the man stretching and applying a force on both sides? Or is he applying a force of 94.9N on each side? Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted May 3, 2006 Report Share Posted May 3, 2006 You appear to have a good understanding of the maths involved :phones:1. In this case the energy stored in the bow can be equated to work done2. I would divide the force by two to get the force for a single hand. Quote Link to comment Share on other sites More sharing options...
kingwinner Posted May 3, 2006 Author Report Share Posted May 3, 2006 You appear to have a good understanding of the maths involved :phones:1. In this case the energy stored in the bow can be equated to work done2. I would divide the force by two to get the force for a single hand.1) Is elastic potential energy = work done all the time? 2) If one side is attached to a wall, he would need 94.9N to stretch the device to 80.0cm apart, and the wall exerts a force of 94.9N on the device in the opposite direction to sum up a net force of zero. Well, if the wall changes back to a hand, would that mean he is exerting a force of 94.9N on EACH hand?...But, wait a second...he is pulling with both hands here, does that mean he can exert half that force on each hand to stretch it to 80.0cm apart? I don't know... The diagram shows 4 forces acting to the right only (F1,F2,F3,F4), is that meaning he is stretching the device with one hand only while keeping the other hand stationary? Quote Link to comment Share on other sites More sharing options...
kingwinner Posted May 4, 2006 Author Report Share Posted May 4, 2006 3) [i am ok with part a, I got an answer of 4.03x10^(-3) m. The hard stuff is part b, with 3 springs. For part b, would the 2 springs (replaced for the rope) have the same spring constant as each other? Are we supposed to find the spring constant for EACH of the 2 springs or 2 of them in total? Would the amount of stretch for the bottom spring be the same as the answer in part a or will it be different?] 4) At an outdoor market, a bunch of bananas is set into oscillatory motion with an amplitude of 20.0cm on a spring with a spring constant of 16.0N/m. It is observed that the maximum speed of te bunch of bananas is 40.0cm/s. What is the weight of the bananas in newtons? (answer: 39.2N)[i am not sure of how to set up this problem. I don't really understand what's happening, what actually is the scenario for this question? Is this spring-mass system horizontal or vertical? Would the equilbrium position change after the bananas are hung? At which point will maximum speed be attained? I am completely lost...] Could somebody explain? Thank you very much! :phones: Quote Link to comment Share on other sites More sharing options...
kingwinner Posted May 4, 2006 Author Report Share Posted May 4, 2006 3) I have assumed everything to be the same for the 2 springs and got an answer of 2.61x10^4 N/m. But I wonder if these assumptions are valid? Quote Link to comment Share on other sites More sharing options...
Qfwfq Posted May 4, 2006 Report Share Posted May 4, 2006 2. I would divide the force by two to get the force for a single hand.:hihi:(rap across the knuckles :confused:) Third law of motion... Quote Link to comment Share on other sites More sharing options...
Jay-qu Posted May 4, 2006 Report Share Posted May 4, 2006 lol, ok I have been told - but I fail to see how the third law anything to do with it... Quote Link to comment Share on other sites More sharing options...
kingwinner Posted May 4, 2006 Author Report Share Posted May 4, 2006 5) Find the spring constant of a spring compressed 0.010m between two cars, each has a mass of 2.5kg, that will cause both cars to move at 3.0m/s. Are the cars actually separating (in oppposite direction) at 3.0m/s or moving together (same direction) at 3.0m/s. Or does that matter? If I assumed the question intended to ask for the first case "separating"Ek=(1/2)mv^2=(1/2)(2.5)(3)^2=11.25J Ee=(1/2)kx^2=(1/2)k(0.010)^2=5x10^(-5) k By law of conservation of energyEe=EkAnd k can be solved... Now, do I have to double Ek to 22.5J? I recall about replaceing one side as a wall. If I replace one car with a wall, the Ek is simply 11.25J, I don't have to double it...but I am not sure about the case of 2 cars... Can someone help me, please? It has been a constant battle in my mind of whether to double the Ek or not. It seems to have a 50% chance of being right in either case... Quote Link to comment Share on other sites More sharing options...
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