Recent content by rmb16

Thankyou! So I fixed my mistake by doing 3.05+X= (1/2)(3.817)(t^2) where x is the distance that the bottom block traveled and its equation is: x=(1/2)(.598427)(t^2) ( I got .598427 by doing 4.7575= 7.95a where 4.7575 is the friction force acting on the bottom block by the top block) and so I...

Homework Statement A small block of mass 1.74 kg rests on the left edge of a block of length 3.05 m and mass 7.95 kg. The coefficient of kinetic friction between the two blocks is 0.279, and the surface on which the 7.95 kg block rests is frictionless. A constant horizontal force of magnitude...