Question

$\boxed{ \bf \: {solve \: this \: physics \: question}}$

Answer 1

In the given condition, The force or frictional force working b/w A & B
That is = coefficient of friction × normal reaction through A

If $m_{a}$ be the mass of the block, then
$f_{ab}$ = μ$m _{a}$ g 

Now for the frictional force b/w B & C 

$f_{bc}$ = μ ($m_{a} + m_{b}$ ) x g.

So frictional force throughout C .
$f_{c}$ = μ($m_{a} + m_{b} + m_{c}$ ) x g.

Now force for a B block , Tension T 

= $f_{ab} + f _{bc}$ 
= μ$m_{a} g$ + μ ($m_{a} + m_{b}$ ) x g
= μ (2$m_{A}$ +$m_{b} . g$

Force along the C block :-

F = $f_{b} + f_{c} + T$
F = μ ($m_{a} + m_{b}$  . g) + μ ($m_{a} + m_{b} + m_{c}$ ) g + μ ($2m_{a} +m_{b}$ ) x g.

Sum of this all
F = μ [ $4m_{a} + 3m_{b} +m_{c}$ ] x g 
Put all the given values of m_{a} , m_{b} & m_{c}  & μ  then 
F = 0.25 ( 4× 3 + 3 × 4 + 8 ) x 9.8
F = 0.25 × 32 × 9.8
F = 0.25 × 313.6
F = 78.4 
Hence the force necessary to drag C at a constant speed is 78.4 N .


Hope it Helps.

Answer 2

Hey there !

It's the above attachment !

Thank you