Question

A body of mass 5kg is placed on a horizontal table having u = 0.4. The mass of 10kg is arranged to move
vertically. The string connecting the two masses is inextensible and the pulley over which the string is
passed is smooth. The acceleration of the system is (g-10 ms 2)​

Answer 1

acceleration of system is 5.33 Meter per second^2

Answer 2

Given:

A body of mass 5kg is placed on a horizontal table having u = 0.4. The mass of 10kg is arranged to move  vertically. The string connecting the two masses is inextensible and the pulley over which the string is  passed is smooth.

To Find:

The acceleration of the system is (g=10 m/s^2)​?

Solution:

it is given that-

mass of 1st body , M1=5kg,

mass of 2nd body , M2=10kg

coefficient of friction , μ = 0.4

take  , g = 10m/s^2

therefore, according to given question,

for block of  mass M1 placed on a horizontal table having μ = 0.4 ,      $M_{1}a=T-\mu M_{1}g \ \ \ ....(1)$  

where, a is the acceleration of the system and T is the tension produced.

for  block of mass M2,      $M_{2}a=M_{2}g-T \ \ ....(2)$

therefore,

On adding equation (1) and (2) , we get

$a(M_{1}+M{2})= g(M_{2}-\mu M{1})\\\\\Rightarrow \ a = \dfrac{M_{2}-\mu M{1}}{M{1}+M_{2}} \ g$

on substituting the given values , we have

$a=\dfrac{10-0.4\times5}{10+5} \ (10)\\\\\Rightarrow \ a=\dfrac{16}{3}$

Hence, acceleration of the system is 16/3 m/s^2.