can you explain where I went wrong and detail information on how to this problem

Question

can you explain where I went wrong and detail information on how to this problem thank you

2. Two blocks connected by a string are pulled across a horizontal surface by a force applied to one of the blocks, as shown. The coefficient of kinetic friction between the blocks and the surface is 0.25. If each block has an acceleration of 2.0 m/s to the right, what is the magnitude F of the applied force? For 3 kg block For the block of 1 Kg Frictional Force- umg = 0.25*1*9.8 Tension = 425N orce on x axis = FCos(60) = 0.5F F sin 60 -2.45 N Let T be the Tension in the String Tymg = ma T-2.45N = (1.0 kg)(2 m/s2) T-2.45N 2N T-2.45N 2N T30k LO Frictional Forceumg+ FSin 60) F cos 60 (-0.25) (3.0) (9.8) + 0.866 7.35 0.866F 425 + 425 + 0.5F-7.35 + 0.866F = ma 842.65 + 1.366F-(4) (2.0) 842.65 +1.366F-6 friction force friction force +2.45N 2.45N = 425N

Explanation / Answer

From free body diagram of 1 kg block

Let tension in string be T

Friction force on block = 0.25*1*9.8=2.45 N(towards left)

T-2.45=ma=2*1=2

T=2.45+2=4.25N

Let us consider 3 kg block:

Applied force=F

Horizontak force = Fcos60=0.5F

Vertical force=F*sin60=0.866F

Normal force on block from ground = 3*9.8 - 0.866F = 29.4-0.866F

Frictional force on block = 0.25*(29.4-0.866F)=7.35-0.2165F

Therefore, for the 3 kg block, 0.5F-4.25-(7.35-0.2165F)=3*2

Or, 0.7165F-4.25-7.35=6

F=24.56 N

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