1. The problem statement, all variables and given/known data A block of mass m1 is placed on an inclined plane with slope angle a and is connected to a second hanging block of mass m2 by a cord passing over a small frictionless pulley the coefficient of kinetic friction is uk the system is released from rest with block m2 dropping a distance h. given m1, a, h, m2, uk calc the work due to each of the forces on block 1 in moving h calc the work due to each of the forces on block 2 in moving h calc the total work on the sys and final speed of the blocks 2. Relevant equations w = F*d 3. The attempt at a solution http://i.imgur.com/IxtPw.png for block1 Wg = -h*mg*sin a Wt1 = hT Wn=0 Wff = -h*uk*mg*cos a Wtot = hT - *mg*sin a - h*uk*mg*cos a for block2 Wg = mgh Wt = -hT Wtot = mgh - hT hT - *mg*sin a - h*uk*mg*cos a + mgh - hT = 1/2*m*Vf^2 + 1/2*m2*Vf^2 - 0 (start from rest) sqrt( 2/(m1+m2) * (mg*sin a - h*uk*mg*cos a + mgh)) = Vf ---- blocks a, b, c are connected as shown. block a,b,c have same mass m and coefficient of kinetic friction between each block and the surface is uk. block c descends with constant velocity. use 30 for angle of incline. given m h uk determine work due to forces on A moving h work due to forces on B moving h work due to forces on C moving h solve for uk in terms of the other givens for the system to move at constant speed http://i.imgur.com/DtwxJ.png block A Wff= -uk*mg*h Wt= T1*h block B Wt2 = T2*h Wg = -mg*sin a*h Wff= -uk*mg*cos a * h Wt1 = -T1*h wtot = T2*h - mg*sin a * h - uk*mg*cos a * h - T1* h block C Wg = mg*h Wt2 = -T2*h Wtot = mg*h - T2*h T1*h - uk*mg*h + T2*h - mg*sin a * h - uk* mg* cos a * h - T1*h + mg*h - T2*h = 0(constant speed implies vf v0 are the same meaning no change in kinetic energy right??) -uk*mg*h - uk*mg*cos a * h = mg*sin a * h - mg*h -uk - uk*cos a = sin a - 1 -uk(1 + cos a) = sin a - 1 uk = - (sin a - 1/ (1+cos a)) uk = 1/2 / (2+sqrt(3))/2 uk = 1/(2+sqrt(3)) want to be sure im not doing anything horribly wrong thanks for any help.