A 171 g wood block is firmly attached to a very light horizontal spring, as shown in the figure below.(adsbygoogle = window.adsbygoogle || []).push({});

The block can slide along a table where the coefficient of friction is 0.306. A force of 20.9 N compresses the spring 17.2 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch on its first swing?

i know the equation is suppose to be 1/2k(x^2-D^2)=(Uk)mg(x+D)

where D is the distance the block will travel....

and then set the equation equals to D = x - (2Ukmg)/D

i plug in all the numbers and i still get the wrong answer...

i think my K constant is wrong i thought it was just k = 20.9 N/.172 m = 121.51

btw my wrong answer is 8.7596 m

can someone help me?

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# Homework Help: Spring and wood block problem

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