STD 11 - 4.1 , newtons laws of motion — JEE STD 11 Science — Question

We have,

$f _1=20\,N , f _2=10\,N$

$m_1=2\,kg , m_2=4\,kg$

Let tension between the blocks is $T$

Here, $F_1 > F_2$

So, the net force acts in the right direction.

for block of mass $2 kg$,

$T - F _2=2 a$

$T-10=2 a \ldots \ldots \ldots(1)$

For the block of mass $4\,kg$,

$F_1-T=4 a$

$20-T=4 a .$

BY solving the above two equation the we get,

$10=6 a$

$\Rightarrow a=\frac{5}{3} m / s ^2$

Thus,

From equation $(1)$ we get,

$T=10+\frac{10}{3}$

$=\frac{40}{3} \,N$