Physics, asked by dannamalaid, 11 months ago

In the given figure shown (see the attachment) , two blocks, one of mass 5kg and the other of mass 2kg are connected by a light and inextensible string. Pulleys are light and frictionless. Find acceleration of both the bodies as well as the tension in the string.

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Answers

Answered by Yeshwanth1245
7

Given▶

m1 = 2kg

m2 = 4kg

g = 10

Let T be the tension in the string due to the two masses!

we have!

weight of 2kg block < Tension in the string!

this can be represented as

T - m1(g) = ==(1)

SIMILARLY,

weight of 4kg block is greater than Tension in the string , it is given by!

m2(g) - T = =(2)

Note!!

____________

We are subtracting the tension from the weight because both are acting in opposite direction i.e, weight acting in downwards , while Tension acting along the length!.

____________

subtracting equation (1)&(2)

T - m1(g) =  

-T + m2(g) =  

we get!

g(m2 - m1) = a(m2+m1)

10(4-2) = a (4+2)

= a

a =  

a = 3.33

substituting values of 'a' in equation (1), we get!

T = m1(g)+ m1(a)

T = m1(g+a)

= 2(10+3.33)

=  

= 26.66N is the tension in the string!

and the acceleration of the masses is

3.33

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Answered by hemakumar0116
1

Answer:

3.33

Explanation:

Given : two blocks, one of mass 5kg and the other of mass 2kg are connected by a light and inextensible string. Pulleys are light and frictionless.

To find : Find acceleration of both the bodies as well as the tension in the string.

Solution :

m1 = 2kg

m2 = 4kg

g = 10

Let T be the tension in the string due to the two masses!

we have!

weight of 2kg block < Tension in the string!

this can be represented as

T - m1(g) = ==(1)

SIMILARLY,

weight of 4kg block is greater than Tension in the string , it is given by!

m2(g) - T = =(2)

We are subtracting the tension from the weight because both are acting in opposite direction i.e, weight acting in downwards , while Tension acting along the length!.

subtracting equation (1)&(2)

T - m1(g) =  

T + m2(g) =  

we get!

g(m2 - m1) = a(m2+m1)

10(4-2) = a (4+2)

a = 3.33

substituting values of 'a' in equation (1), we get!

T = m1(g)+ m1(a)

T = m1(g+a)

= 2(10+3.33)

= 26.66N is the tension in the string!

and the acceleration of the masses is 3.33

=3.33

=3.33 Is the correct answer of this question .

#SPJ3

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