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Question

Block A has a weight of 300N and block B has a weight of 50 N. If the coefficient of kinetic friction between the incline and block A is μk = 0.2. Determine the speed of block A after it moves 1m down the plane, starting from rest. Neglect the mass of the cord and pulleys.
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Solution

R.E.F image
N=300cosθ
=300×45=240N
Let us take,
Velocity of block be A after it moves 1 m
down the plane be V
and velocity of block B be VB
We know that,
in pulley system
T.U = constant
27.u=πuB
uB2u
and T.x= constant
27×1=f2
x=2m
Since N=240N
f=uN=0.2×240
f=48N
Applying energy conservation
Decrease in potential energy of block A
= Increase in kinetic energy of block A
+ Increase in kinetic energy of block B
+ Increase in potential energy of block B
+ work done by friction force
300×1sinθ=12×300g×v2+12×50g(2v)2+50×2+48×1
300×35=15v2+10v2+100+48
=180=25v2+148
v2=18014825
v2=3225
v=425

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