1. The problem statement, all variables and given/known data A 20.0 kg block on a horizontal surface is attached to a horizontal spring of k = 2.0 kN/m. The block is pulled to the right so that the spring is extended 10.0 cm beyond its unstretched length, and the block is then released from rest. The frictional force between the sliding block and the surface has magnitude 80.0 N. 2. Relevant equations What is the maximum KE attained by the block as it slides from release to the point where the spring is unstretched? 3. The attempt at a solution KE is maximized when velocity is maximized. Therefore, the point is when the two forces are at equilibrium. Set kx = Fd, d = (0.1-x) 2000(x) = 80(0.1-x) x = 8/2080 (m) = 0.003846 (m) EPE (when fully stretched to 0.1 m) + Work done by friction = KE + EPE2 (.5)(2000)(0.1^2) - 80(0.1-0.003846) = .5(2000)(0.003846^2) + KE Therefore, KE = 2.29 J, which makes sense. I also did it the calculus way W = the integral of force F = ma, m =20 kg a = [kx-f(0.1-x)]/20, integrating this yields velocity function, c =0 derive that function and set it to zero, yields the same answer. But 2.29 J is not right. Why?