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# Hooke's Law

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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/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?

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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 done

2. I would divide the force by two to get the force for a single hand.

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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 done

2. 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?

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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:

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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?

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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...

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lol, ok I have been told - but I fail to see how the third law anything to do with it...

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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 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...

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