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anonymous
 one year ago
Classical Mechanics Series; Hooke's Law
anonymous
 one year ago
Classical Mechanics Series; Hooke's Law

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0This is a simple tutoring on Hooke's law.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Hooke's law identifies the constant of a given string... or k. k=F(applied)/Delta x Where k denotes the force required to stretch or compress a spring per unit of length.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Usually K is measured in N/m. Let's keep that as our rule of thumb in solving the following question. The length of a stretched spring is 2m with an applied force of 16N. To identify the k constant for the spring, K=16N/2m=8N/m Therefore the k constant for this spring is 8N/m

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0p.s Note that there is also a reference point in spring as well. You would measure the stretch or compression from the point of zero.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0Alright, the next equation concerning Hooke's law is this, E(energy stored in a compressed or stretched spring)=1/2(k)Delta(x) Where, E=potential stored k=Hooke's constant x=stretch, or compression of the spring. With this formula alone you can calculate the energy stored in a spring, or potential energy from a spring being stretched or compressed. Therefore, you look at the problem Calculate the spring constant of a spring if a 15 kg mass is suspended from the spring and it stabilizes with a stretch of 12 cm. First you need to look at the resulting force acting downwards to the gravity source, which is accounted for by (9.8m/s^2)(15kg)=147N Next step is you must convert the 12cm to meters 12cm/100cm/m=0.12m Then you would calculate the k constant L=147N/0.12m=1225N/m This tell us how much force is required to stretch or compress the spring by a meter. Now using the formula E=1/2kx^2 E(potential energy stored in the spring)=1/2(1225N/m)(0.12cm)^2=8.82J After accounting for sig figs of 2, The energy stored in the spring is 8.8J.

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0Neat..... Nice work robert! Tell us about Kinetic energy also... :)

arindameducationusc
 one year ago
Best ResponseYou've already chosen the best response.0I will be referring this tutorial in my tutorial, @Robert136
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