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# How much coins weight would it take to stretch a band by one

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(a) Based on the linear fit equation y=395.7x + 1.44 (this equation is derived from a graph where stretch of a band in meters is on the x axis and weight is on the y axis), how much coin weight would it take to stretch the band by one meter?

The force constant k of the band is 395.7.

(b) How does above linear equation relate to Hooke's law F= kx ?

https://brainmass.com/physics/newtons-laws/determining-force-and-weight-coin-example-question-313738

## SOLUTION This solution is FREE courtesy of BrainMass!

(a) I see from your description that in the equation, y = 395.7x + 1.44, x is the distance in m by which the band is stretched and y is the total force (sum of all the coin weights).

To find how much force is required to stretch the band by 1 m, plug in x = 1 into the above function.

y = 395.7(1) + 1.44
= 397.14

(b) F = kx and y = 395.7x + 1.44 are both linear equations and indicate a direct variation of y with x. F = kx means the force of deformation is directly proportional to the stretching distance. Note that just by comparing the two equations, you get k = 395.7, which is your spring constant.

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