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! Please Register or Log in to view the hidden image! 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) Please Register or Log in to view the hidden image! Total spring constant=65.0x4=260N/m Amount of stretch from equilibrium position=0.365m F=kx =(260)(0.365) =94.9N Now 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?
Work is force applied over a distance. Energy is the capacity to do work. By the conservation of energy, as you do work on the bow you increase it's potential to do work (elastic potential energy) by the same amount (neglecting losses). They are not the same thing, but they are typically equal. You know how to do this problem if one of the hands is changed to a wall. Think of the problem that way. Now, how much force does the remaining hand have to exert? Then, given the fact that the hand is giving that much force on one side and the springs are not accelerating how much force is the wall exerting on the other side? Now, change the wall back to a hand. -Dale
Hi, DaleSpam, 1) So is elastic potential energy = work done all the time? And when I use W=Fd(cos theta) for this question, I get an answer of 92J instead of 46J, it's doubled, why? 2) If one side of the device is attached to a wall, he would need 94.9N to stretch the spring over that distance. But in this case, he seems to be stretching the springs in BOTH directions. Does that mean each hand is exerting 47.4N on the device? (i.e. a total stretching force of 94.9N?) But I wonder if that's true...I am confused...
Not sure... Please Register or Log in to view the hidden image! The device is being stretched, so it's not stationary at all (the spring is moving), so I guess the net force on it may not be zero...
OK! 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... Please Register or Log in to view the hidden image! 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?
So to stretch it between the wall and the hand by 80cm you need two equal and opposite forces of 94.9N each. Now, the springs do not know anything about walls and hands so they don't care if that force comes from a wall or a hand, so replacing the wall-force by a second hand force ... -Dale
Oh, I see! So we don't have to divide the force by 2 to get the force exerted by a single hand. That's tricky. At first glance, it seems that we have to divide by 2, but actually we don't!
3) Please Register or Log in to view the hidden image! [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! Please Register or Log in to view the hidden image!
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?
3) Your answer looks fine, and the two springs have to be similar for them to have the same stretch and still be in equilibrium. 4) The system is probably vertical. but it wouldn't matter if it were horizontal (assuming the spring resisted compression and stretching the same way). Gravity would offset the entire oscillation the same way, so you can ignore it.
3) So the amount of stretch of the bottom spring is different than that of #3a...For #3b, is it possible to determine the amount of stretch of each of the 3 springs? It seems that there is an infinite number of such values... 4) How can I apply the law of conservation of energy to solve this problem? Which 2 points should I use for calculations? And if it's horizontal, how can we determine the weight? (it's not the force causing the oscillation if the system is horizontal). If the spring-mass system is vertical and frictionless, will the oscillation be constant (having the same amplitude)? (i.e. moving up and down to the same height for each cycle?) Or will it stop oscillating completely over time?
4) When I assume the spring is horizontal, I got the correct answer. But I don't know how to do it when it's vertical, becuase with the mass hanging, the new equilbrium position and the unstreched position are NOT the same...
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 energy Ee=Ek And k can be solved... Now, do I have to double Ek to 22.5J? I recall DaleSpam hinting 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...
Yes, exactly. Sorry about dragging you through the reasoning process step by step, but you will undoubtedly run into similar problems elsewhere. Hopefully the next time you see a similar tricky question you will be able to piece it together even if the exact situation is slightly different. -Dale
Thanks DaleSpam! I ran into a similar problem for question 5, but then I can't decide whether to multiply the kinetic energy by 2 or not... Please Register or Log in to view the hidden image!
I read "compressed between two cars" as indicating that the cars wind up travelling in opposite directions, but honestly it does not matter in this case. The problem is explicit that each car has a mass of 2.5 kg and each car winds up with a velocity of 3.0 m/s. That means that each car has a KE of 11.25 J so the total system has a KE of 22.5 J. There is no confusion about one or two in this case since the problem is explicit. -Dale
So does the spring get detached from both cars after returning to its unstretched position with the 2 cars moving in opposite directions? I see, so the ealstic potential energy required to move both cars at 3.0m/s^2 is 22.5J. When I replace one car with a wall, and compress the car against the wall, the Ek is just 11.25J, I don't have to double it, so why do I have to double it if the wall is replaced back to the car? And I was thinking about action and reaction forces that seems to suggest 11.25J will be enough.....I was like completely trapped by these little tricky stucking points.... Please Register or Log in to view the hidden image!
Excellent point. Let's think about that situation a bit. As we already saw before the force on the wall is the same as the force on the car*. However, work is not just force but force times distance. Despite the force exerted on the wall the wall does not move so the work done on the wall is 0 (corresponding to the final KE of 0 for the wall). With a wall there all of the spring expansion works at moving the car, while with two cars some of the spring expansion works at moving the other car. -Dale *technically that is only true in general for a spring at rest, but if the spring is approximately massless then it is approximately true for a moving spring as well
