assume that block on the table has twice the inertia of the

assume that block on the table has twice the inertia of the hanging block (string attached to the block and hanging block and pulley on the edge of table to hang the hanging block).

A.You give the block on the table a push to the right so that it starts to move. If the magnitude of the frictional force exerted by the table on the blokc is half the magnitude of the gravitation force exerted on the hanging block, what is the acceleration of the hanging block after you have stopped pushing the block on the table? B. What woudl this hanging block acceleration be if you had pushed the block on the table to the left intead? consider only a short time interval after you stop pushing

Solution

(A)
Let mass of block on tabel = 2m
Hanging Block mass = m
Let coefficient of Friction = u

Fr = u*2m*g
Grabitational Force , Fg = m*g

Given,
2* Fr = Fg
2* u*2m*g = m*g
u = 1/4 = 0.25

Let the acceleration = a in downwards direction
Fg -T = m*a ----------1

T - Fr = 2m*a ------------2

Using equation 1 & 2 -
Fg - (2m*a + Fr) = m*a
Fg - Fr = 3m*a
2*Fr - Fr = 3*m*a
Fr =  3*m*a
u*2m*g = 3*m*a
a = 2*0.25*9.8/3
a =  1.63 m/s^2

(B)
In this Case both the Friction Force and Tension in pulley cable due to hanging weight would add up in the same direction towards right and would cause acceleration.

T = m*g
Friction Force = u*2m*g

Total Tension in the pulley cable =  m*g+ 2*u*m*g
Total Tension = 1.5 m*g
Acceleration, a = 1.5m*g/2*m
a = 1.5*9.8/2
a = 7.35 m/s^2

Initial Acceleration of the block would be = 7.35 m/s^2