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Two blocks, m1 = 10 kg and m2 = 5 kg, are joined by a massless inextensible string of length 0.3 m and laid along a diameter of a turntable. Friction coefficient between m1 and the table is 0.5; m2 is frictionless on the table. The table rotates at 10 rad/s about a vertical axis through its centre O, with the masses on opposite sides of O and m1 at 0.124 m from O. The blocks remain at rest relative to the table. Find the frictional force acting on m1.

1. 28 N
2. 32 N
3. 36 N
4. 40 N

Correct answer: 36 N

Solution

m2 is at radius 0.3 - 0.124 = 0.176 m; tension T = m2*w²*r2 = 5*100*0.176 = 88 N. For m1, required centripetal = m1*w²*r1 = 10*100*0.124 = 124 N, supplied by friction + tension, so f = 124 - 88 = 36 N.

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