Determine the acceleration of the `5 kg` block `A` . Neglect the mass of the pulley and cords . The block `B` has a mass of `10 kg`, The coefficient of kinetic friction between block `B` and the surface is `mu _(k) = 0.1` (take `g = 10 m//s^(2))`

Explanation

Soppose `T` be the tension in the string attached to block `B` . Then tension in the string connected to block `A` would be `4 T`

Similarly, if a be the acceleration of block `B` towards right will be `4 a`
Equation of motion are
For block `A , m_(A) g - 4 T = m_(A) a`
or ` 50 - 4 T = 5 a` ...(i)
For block `B T - f = 10 (4 a)`
or `T- (0.1) (10) (10) = 40 a`
or `T- 10 = 40 a`
Solving Eqs, (i) and (ii), we get
`a = (2)/(33) m//s^(2)`