Question

A mass of 2m sits on a table and a mass-less cord runs from it over a frictionless pulley also with negligible
mass. Attached to the hanging cord is a mass of 3m. If the u = 0.5, the tension in the cord is
=5
(A) 3mg
(B) 2mg
(C) 5mg/2
(D) 9mg/5​

Answer 1

Given:

A mass of 2m sits on a table and a mass-less cord runs from it over a frictionless pulley also with negligible mass. Attached to the hanging cord is a mass of 3m.

To find:

Tension in the cord ?

Calculation:

Let the tension in the cord be T :

Now , for 3m block:

$\therefore \: (3m)g - T = (3m)a \: \: \: .....(1)$

Now, for 2m block, the friction will act opposite to the tension :

$\therefore \: T - \mu \{(2m)g \} = (2m)a \: \: \: .....(2)$

Adding eq.(1) and eq.(2):

$\therefore \: 3mg - 2\mu mg = 3ma + 2ma$

$\implies \: 3mg - 2\mu mg =5ma$

$\implies \: 3mg - 2(0.5) mg =5ma$

$\implies \: 3mg - mg =5ma$

$\implies \: 2mg =5ma$

$\implies \: ma = \dfrac{2}{5} mg\: \: \: .....(3)$

Now, putting value of (ma) in eq.(1):

$\therefore \: (3m)g - T = (3m)a$

$\implies\: 3mg - T = 3( \dfrac{2}{5} mg)$

$\implies\: 3mg - T = \dfrac{6}{5} mg$

$\implies\: T =3mg - \dfrac{6}{5} mg$

$\implies\: T = \dfrac{15 - 6}{5} mg$

$\implies\: T = \dfrac{9}{5} mg$

So, final answer is:

$\boxed{ \bold{\: T = \dfrac{9}{5} mg}}$

Answer 2

Explanation:

D}. 9mg/5

I hope it may help to you..