Question

Three masses of 1kg,6kg and 3kg are connected to each other with threads and are placed on the table as shown in figure. What is the acceleration with which the system is moving?

Answer 1

Force= mass * acceleration
mass of system= 1+6+3= 10kg
But, Force = mass * g
Force= 10*9.8
Force = 98N

Now,
F = ma
98= 10*a
a=98/10
a = 9.8m/s^2

I've assumed the value of g as 9.8m/s^2.
The answer may be wrong as the figure isn't shown by you.

Answer 2

Given :-

${m}_{1}$ = 1 kg

${m}_{2}$ = 6 kg

${m}_{3}$ = 3 kg

For 1 Kg block,

${T}_{1}-{m}_{1}g = ma$

${T}_{1}$ - 10 = a -------(1)

For 6 Kg block,

${T}_{2}-{T}_{1}$+ 60 = 6a -----(2)

For 3 Kg block,

30 -${T}_{2}$ = 3a ----(3)

On adding equation (1), (2) and (3) we get,

20 = 10a

a = 20/10

a = 2 ms-².