Two masses m_{1} = 5 kg and m_{2} = 10 kg, connected by an inextensible string over a frictionless pulley, are moving as shown in the figure.

The coefficient of friction of horizontal surface is 0.15. The minimum weight m that should be put on top of m_{2} to stop the motion is

- 23.3 kg
- 43.3 kg
- 10.3 kg
- 18.3 kg

**Answer**

m_{1} = 5 kg and m_{2} = 10 kg

As the system is at rest, T = 50 N

50 – T = 5 × a

T – 0.15(m + 10) g = (10 + m)a

a = 0 for rest

50 = 0.15(m + 10) × 10

5 = 3/20(m + 10)

100/3 = m + 10

m = 23.3 kg

**The correct option is A.**