Physics, asked by kalpana9170, 4 months ago

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​

Answers

Answered by nirman95
2

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}}

Answered by Anonymous
1

Explanation:

D}. 9mg/5

I hope it may help to you..

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