A wood block with a mass of 100 grams is held at the top of an inclined board (point A in Figure 2). If we assume force of friction (Ff) is 0.35N and constant as the block slides down the ramp, what will the velocity be at point B after the block is released and slides down the surface?
ok, the component of gravity down the plane is .1*9.8*cosTheta. That is the force down the plane.
net force=m*a
.98cosTheta-.35=.1*a
solve for acceleration a.
Vf^2=2*a*distance.