Recent content by APPhysic

Well, for the x component, ƩFx=T-fk=m1a1 And for y, ƩFy=n-m1g So then, can I say that T-μkm1g=m1a1 ? m1 in this case being the aluminum block

ehild, Thanks for the welcome! I believe the aluminum block has tension, normal & gravitational force, and the force of friction acting on it. And won't the copper and aluminum block have the same acceleration or no?

Homework Statement A 1.00-kg aluminum block and a 4.00-kg copper block are connected by a light string over a frictionless pulley. The two blocks are allowed to move on a fixed steel block wedge (of angle 31.0°). The aluminum block is on the left side, on a flat plane. On the right is the...