1. The problem statement, all variables and given/known data A block of mass 10kg moves along a surface inclined 30° to the horizontal. The centre of gravity of the block is elevated by 3m and the kinetic energy of the block decreases by 50J. The block is acted upon by a constant force Rparallel to the incline and by the force of gravity. Assume frictionless surfaces and let g= 9.81 m/s**2. Determine the magnitude and direction of the constant force R. 2. Relevant equations ΔK = m*(Vb^2 - Va^2)*(1/2) , ΔU=mgz 3. The attempt at a solution Since i am unable to express mathëmatically the ""The centre of gravity of the block is elevated by 3m and the kinetic energy of the block decreases by 50J"" i will try to ignore it for now. The force R can be either applied to the left or the right,we don't know yet.I will assume that it is towards the right,if not then the sign will be minus in my final calculations. -- ΣFy = 0 , N=Wy=mgcos30° ( the COM being the origin and x axis parallel to the inclined surface) -- ΣFx = Wx - R i) How do we even change the COM of a block ? ii)Why is the kinetic energy lost ( 50J)not equal to the potential energy gained ? In other words 50 <> m*g*( (h+3)-h) which makes sense mathematically but since energy is conserved where does it go ? I am really confused guys !