Please show work:$-$

The horizontal surface on which the objects slide is frictionless. If $M=2\mathrm{.}0kg,$ the tension in string $1$ is $12$ N$.$ Determine $F.$

$F=(M+2M+2M)a=5Ma(1)$

for ist $2$ :

$F-12=2Ma(2)$

a $25N$ from $(1)ma=\frac{F}{5},$ then $(2)$ becemes

c$.30N$

d$.35N$

$F-12=2(\frac{F}{5})5F-60=2F$

e$.40N$

$3F=60NF=20N$

ANS: B

PTS: $2$

DIF: Average