Question

in the arrangement of figure assume negligible friction between the blocks and table.If F the pulling force m1 and m2 the masses are known then the tension in the string is

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