Question

Two masses m₁ = 5kg and m₂ = 10kg, 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) 27.3 kg 

(2) 43.3 kg 

(3) 10.3 kg 

(4) 18.3 kg

Answer 1

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

Answer 2

m₁g = μ(m₂ + m)
5 = 0.15 (10 + m)
500/15 = 10 + m
m = 350/15 = 23.3

So, correct answer will be option (1)