# Physics

A horizontal spring(k= 86.2 N/m) is placed on a countertop and is compressed a distance of 36.3 cm. A 103.5 g object is placed at the end of the spring. The coefficient of kinetic friction between the object and countertop is measured to be .47. What acceleration will the object experience when the spring has moved a distance of 13.2 cm back to its rest state?
Could you please all the steps? I don't know where to start.

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1. M*g = 0.1035 * 9.8 = 1.014 N. = Wt. of object. = Normal force(Fn).

F = 86.2 N/m * 0.132m = 11.4 N.

Fk = u*Fn = 0.47 * 1.014 = 0.477 N. = Force of kinetic friction.

F-Fk = M*a.
11.4 - 0.477 = 0.1035a, a = ?.

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