Question

In the arrangement of figure assume negligible friction between the
blocks and table. If F the pulling force and m, and m, the masses are
known, then the tension in the string is
W25 kg
(A) m,F/ (m, + m2)
2m F/ (4m, + m)
(B) 2m, F / (m, + m)
(D) None of these​

Answer 1

Answer:

$\frac{m_1F}{m_1+4m_2}$

Explanation:

Let the tension in the string is T

If acceleration of mass m1 is a then the acceleration of mass m2 will be 2a

$T=m_2\times 2a$

$\implies T=2m_2a$

And

$F-2T=m_1a$

$\implies F-2(2m_2a)=m_1a$

$\implies F-4m_2a=m_1a$

$\implies F=(m_1+4m_2)a$

$\implies a=\frac{F}{m_1+4m_2}$

Therefore Tension T

$T=m_1\times\frac{F}{m_1+4m_2}$

$\implies T=\frac{m_1F}{m_1+4m_2}$

Hope this is helpful