A 6.00 kg block is set into motion up an inclined plane with an initial speed of v0 = 8.50 m/s (Fig. P7.23). The block comes to rest after traveling 3.00 m along the plane, which is inclined at an angle of 30.0° to the horizontal. (a) Determine the friction force exerted on the block (assumed to be constant). (b) What is the coefficient of kinetic friction? Wfk + Wg = -.5mvi2 (fk)(x) + (m)(g)(h) = (-.5)(m)(vi2) (fk)(3) + (6)(9.8)(3sin30) = (-.5)(6)(8.5^2) fk = -101.65 Final answer is incorrect. Cannot move on to (b) without understanding (a). Any ideas?