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oreosama
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Homework Statement
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
Homework Equations
w = F*d
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
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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 I am not doing anything horribly wrong thanks for any help.