Two masses m1 = 5 kg and m2 = 10 kg, connected by an inextensible string

• Laws of Motion

Two masses m₁ = 5 kg and m₂ = 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 m₂ to stop the motion is

1. 23.3 kg
2. 43.3 kg
3. 10.3 kg
4. 18.3 kg

Answer

m₁ = 5 kg and m₂ = 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.