Question

A block A of mass m1 rests on a horizontal table. A light string connected to it passes over a frictionless pulley at the edge of table and from its other end another block B of mass m2 is suspended. The coefficient of kinetic friction between the block and the table is µk​. When the block A is sliding on the table, the tension in the string is
​
i am getting an ans:g(m1m2(uk+1)+​​ukm2^2) please tell me hoe to get the orginal ans thank you
ans
m1m2(l+µk )g/(m1 +m2)

Answer 1

oh it's okay , Let's try to solve !

question said , a block 'A' of mass m1 rest on horizontal table and a light string connected it over frictionless pulley at the edge of the table from it other end another block of mass m2 is suspended.

when block is sliding , kinetic friction acts on contacting surface of block A and surface of the table as shown in figure.
so, we can write equation ,
$T-f_k=m_1a$ .......(i)

similarly, for block B we can write equation,
$m_2g-T=m_2a$.......(ii)

from equation (i) and (ii), we get,
$m_2g-f_k=m_1a+m_2a\\\\m_2g-\mu_km_1g=(m_1+m_2)a\\\\a=\frac{(m_2-\mu_km_1)g}{m_1+m_2}$

now put a in equation (ii),
$m_2g-T=m_2\frac{(m_2-\mu_km_1)g}{m_1+m_2}\\\\T=m_2g-m_2\frac{(m_2-\mu_km_1)g}{m_1+m_2}\\\\T=\frac{m_1m_2g+m_2^2g-m_2^2g+\mu_km_1m_2g}{m_1+m_2}\\\\T=\frac{m_1m_2(1+\mu_k)g}{m_1+m_2}$

Answer 2

Hope it helps! Use free body diagram for both masses m1 and m2. Derive equations from them and solve as given.