A 171 g wood block is firmly attached to a very light horizontal spring, as shown in the figure below. 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?