1. The problem statement, all variables and given/known data A 28-kg rock approaches the foot of a hill with a speed of 15 m/s. This hill slopes upward at a constant angle of 40.0∘ above the horizontal. The coefficients of static and kinetic friction between the hill and the rock are 0.75 and 0.20, respectively. a)Use energy conservation to find the maximum height above the foot of the hill reached by the rock. 2. Relevant equations Conservation of energy for this problem is: PEi + KEi + Workstatic friction = PEf + KEf + Workkinetic friction Wfriction = μmgcos(Φ) Friction force = μmg 3. The attempt at a solution m=28kg vi=15m/s Φ=40° μs=0.75 μk=0.20 fs=0.75mgcos(Φ) fk=0.20mgcos(Φ) Wfs=.75mgcos(Φ)*s*cos(180) Wfk=0.20mgcos(Φ)*s*cos(180) plug all known variables into my equation for conservation of energy: (.5)(28)(152)+(28)(9.8)(0)+(0.75)(28)(9.8)(cos(40))*s*(-1) = (.5)(28)(02)+(28)(9.8)(hf)+(0.20)(28)(9.8)(cos(40))*s*(-1) which equals: 3375J - 157.65*s = 274.4*hf - 42.041*s Then i tried to find s which is the displacement vector by doing Newton's second law ΣF= max = wx - fk 28a = 28*9.8*sin(40°) - 0.2*28*9.8*cos(40) 28a = 134.34 a = 4.80m/s2 Then i used vf2 = v22 + 2ad, where d would be my s vector 02 = 152 + 2(4.80)d d = -225/9.6 = -23.43 which i made positive because i just wanted the magnitude. But then my final answer is hf = 22.18m, which isn't correct.