1. The problem statement, all variables and given/known data A 5.0 kg mass is released from rest, at the top of a frictionless 30° inclined plane. Calculate the kinetic energy of the mass when it has slid 40cm down the incline starting from rest. (ANS = 9.80 J) 2. Relevant equations KE=1/2(mv^2) a=v/t d=v1t + 1/2a(t^2) v2^2=v1^2 + 2ad v2=v1 + at d= (v1+v2/2)*t 3. The attempt at a solution G: v1= 0 ; θ= 30° ; d=40cm R: KE=? ; v2=? ; t=? (NE direction is +ve) (1) v2y^2=v1y^2 - 2aydy v2y^2= √0 -2(9.80)(-0.40sin30) v2y= 1.98 m/s (2) v2y=v1y +at t = v2y/ay t = 0.202 s (3) v2x= dx/t v2x= 0.40cos30/0.202 v2x= 1.71 m/s (4) v2= √(1.98)^2 + (1.71)^2 v2= 2.62 m/s (5) KE = 1/2(mv^2) KE = 1/2(5.0*2.62^2) KE = 17.2 J My answer is far off from the correct one so i'm assuming my approach might be wrong.. But I just don't know from where it is that i'm doing it wrong.. Probably something with my calculations, not too sure entirely..