FrictionMedium
Question
Two masses m1 = 4 kg and m2 = 2kg are connected with an inextensible, massless string that passes over a frictionless pulley and through a slit, as shown. The string is vertical on both sides and the string on the left is acted upon by a constant friction force 10 N by the slit as it moves. (use g = 10 m/s2)
Options
A.Acceleration of mass m1 is 5/3 m/s2, downwards.
B.Tension in the string is same throughout.
C.Force exerted by the string on mass m2 is 70/3 N
D.If positions of both the masses are interchanged, then 2kg mass moves up with an acceleration 10/3 m/s2.
Solution
Applying NLM on the part that moves through slit.
T2 − f − T1 = 0
For 4 kg mass 40 − T2 = 4a
For 2 kg mass T1 − 20 = 2a
On solving 10 = 6a
a = 5/3 m/s2.
Force exerted on 2kg mass by string = T1 = N 70/3.
Tension in the string will not be same throughout, due to the friction force exerted by the slit.
T2 − f − T1 = 0
For 4 kg mass 40 − T2 = 4a
For 2 kg mass T1 − 20 = 2a
On solving 10 = 6a
a = 5/3 m/s2.
Force exerted on 2kg mass by string = T1 = N 70/3.
Tension in the string will not be same throughout, due to the friction force exerted by the slit.
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