Question bank/JEE Main/Physics

Two blocks of masses 2 kg and 4 kg are connected through a massless
inextensible string. The
co-efficient of friction between 2 kg block and ground is 0.4 and the
coefficient of friction between 4 kg block and ground is 0.6. Two forces
F1 = 10 N and F2 = 20 N are applied on the blocks as shown in the
figure. Calculate the frictional force ( in N )between 4 kg block and
ground (Assume initially the tension in the string was just zero before
forces F1 and F2 were applied)
[IMAGE] 2 kg 4 kg 2 kg
0.4 4 kg
0.6 F1 = 10 N F2 = 20 N
4 kg (, F1 F2 )

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System will not move (Equilibrium)
For equilibrium of 2 kg block
[IMAGE] 10 = 8 + T
⇒ T = 2N
For equilibrium of 4 kg block
[IMAGE] T + f2 = 20
⇒ f2 = 18 N
TOPIC: NLM
SUB TOPIC: FRICTION

Source: Grass-G / P-03 Paper 1 (Physics) Eng. + Hindi 24-8-2020.docx