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imdumb

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- Thread starter imdumb
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imdumb

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- #2

quasar987

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you have to consider the rubber band has happy Hookian springs. I.e. such that F = kx.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not gonna be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.

By "It takes a force of 15 N to stretch either one of the bands 1.0 cm", what they mean is that if you exert a force of 15N on one of the rubber band, you're not gonna be able to stretch it farther than 1cm. At this point, equilibrium will have been attained. this means that at x = 1cm, the force of the rubber band is of 15N too. From this, you get k = 15 [N/cm].

You go from there.

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- #3

imdumb

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so... then would 7.5 be the correctanswer to part A?

- #4

quasar987

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Nope. How did you get that answer?

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)

(Btw - I fixed some dimension errors in my first post. I replaced all meters with centimeters)

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- #5

imdumb

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I used the equation PE=.5kx^2and put 15 infor k and 1 in for x.

- #6

quasar987

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The actual problem talks about a rock that is pulled back.12 m (12 cm, not 1cm). Moreover, 15 [N/m] is the spring constant of ONE rubber band. You got two.

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