A rope of negligible mass passes over a pulley of negligible mass attached to the ceiling, as shown in figure. One end of the rope is held by Student A of mass 70 kg, who is at rest on the floor. The opposite end of the rope is held by Student B of mass 60 kg, who is suspended at rest above the floor. The minimum acceleration a₀ with which the Student B should climb up the rope to lift the Student A upward off the floor.

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$\frac{1}{3} m / s^{2}$

$\frac{2}{3} m / s^{2}$

$\frac{4}{3} m / s^{2}$

$\frac{5}{3} m / s^{2}$

answer is D.

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Detailed Solution

For student A to just lift off the floor, tension T in string must be greater than or equal to 700 N.
The F.B.D. of student B is
Applying Newton's second law

$\begin{array}{l} T - m g = m a \Rightarrow 700 - 600 = 6 a \\ or a = \frac{5}{3} m / s^{2} \end{array}$

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