Question

Question #51cf4

Answer 1

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`(0.30*N)` or, `(0.30*4mg)` 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*N`=`u*mg`=ma(a is the acceleration of the whole system)
Or,u=`(a/g)`....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)=`3m*a`...2
And for mass 2m we can write,
2mg-T=`2m*a`....3
Solving all three,we get,
a=0.133g
T=1.734g
And u=0.133