JEE Main 2026 — Jan 24, Shift 1 — Physics solution

JEE Main2026Jan 24, Shift 1Physics

In the given figure the blocks A, B and C weigh 4 kg, 6 kg and 8 kg respectively. The co-efficient of sliding friction between any two surfaces is 0.5. The force $F$ required to slide the block C with constant speed is _______ N. (Used $g = 10$ m/s $^{2}$ )

Your answer

Solution

The string-and-pulley arrangement constrains blocks A and B so that they remain effectively stationary while C slides at constant speed. Kinetic friction therefore acts at all three contact surfaces.

Friction values:

Between A and B: $f_{A B} = μ m_{A} g = 0.5 \times 4 \times 10 = 20$ N.

Between B and C: $f_{B C} = μ (m_{A} + m_{B}) g = 0.5 \times 10 \times 10 = 50$ N.

Between C and ground: $f_{gr} = μ (m_{A} + m_{B} + m_{C}) g = 0.5 \times 18 \times 10 = 90$ N.

Tension in the string (from equilibrium of block B): the string must balance both friction forces acting on B:

$T = f_{A B} + f_{B C} = 20 + 50 = 70$ N.

Equilibrium of block C at constant velocity: $F$ balances the ground friction, the friction from B at C's top surface, and the tension transmitted through the string:

$F = f_{gr} + T + f_{B C} = 90 + 70 + 50 = 210$ N.

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