JEE Main | 2018FrictionEasy
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 :-
Options
A.
27.3 kg
B.
43.3 kg
C.
10.3 kg
D.
18.3 kg
Solution
50 T = 5 a
T 0.15 (m + 10) g = (10 + m)a
a = 0 for rest
50 = 0.15 (m + 10) 10
5 = (m + 10)
= m + 10
m = 23.3 kg
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