Question

Did I do a) and b) right?
The three blocks shown are released from rest and are observed to move with accelerations that have a magnitude of 1.5 m/s2. Disregard any pulley mass or friction in the pulley and let M = 2 kg.
a) What is the magnitude of the friction force on the block that slides horizontally?
b) What is the tension force between M and 2M?

pulley---2M---pulley
rope..............rope
M...................2M
massM is hung by a rope under pulley on the left. One mass 2M is in the middle and another mass2M is hung by a rope under right pulley.

For a) this ishow I do =>F=5Ma=> F=5*2*1.5=15N=> F=2Mg-Mg-Ff=> 15N=2*2*9.81-2*9.81-Ff=> -Ff=15-N-19.62N=-4.6N=> Ff=4.6N

For b) this how I do => a=(F+mg)/(2M+M+2M)=> a=(4.6+2*9.81)/(2*2+2+2*2)=1.5m/s^2

T=F2Ma=> 4.6+2*2*1.5=10.6N

Answer 1

I dont follow your work.
a) the total pulling force is 2mg
net force=totalmass*a
2Mg-Ff-Mg=5M*1.5 (the acceleration of the system is given)
Ff=M*9.8-7.5M=2.3M=4.6N

b. tension: the last segment of rope is pulling the last block only.
tension=M(g+a)=2(9.8+1.5)=2(11.3)=22.3 N