# Question #51cf4

Jan 11, 2018

Acceleration=0.133g
Tension in the string=1.734 N
Coefficient of kinetic friction=0.133
Frictional force acting at 3m and table interface is$\left(0.30 \cdot N\right)$ or, $\left(0.30 \cdot 4 m g\right)$ i.e 1.2mg

#### Explanation:

For both of the block of mass m and 3m to move together there must be the presence of frictional force at their interface.
Let's assume the coefficient of kinetic friction at m and 3m interface is u
So,we can say,
Frictional force acting at m and 3m interface(fk1) will be responsible for the forward movement of the mass m along with 3m
I.e fk1=$u \cdot N$=$u \cdot m g$=ma(a is the acceleration of the whole system)
Or,u=$\left(\frac{a}{g}\right)$....1
Now for mass 3m considering free body motion we can say that the tension in the string helps in its forward movement by overcoming the frictional forces acting at m and 3m interface(fk1)& 3m and table interface(fk2),
Hence we can write,
T-(fk1+fk2)=$3 m \cdot a$...2
And for mass 2m we can write,
2mg-T=$2 m \cdot a$....3
Solving all three,we get,
a=0.133g
T=1.734g
And u=0.133