$A$ monkey of mass $40 \, kg$ climbs on a rope which can stand a maximum tension of $600 \, N$. In which of the following cases will the rope break: the monkey
$(a)$ climbs up with an acceleration of $6 \, m \, s^{-2}$
$(b)$ climbs down with an acceleration of $4 \, m \, s^{-2}$
$(c)$ climbs up with a uniform speed of $5 \, m \, s^{-1}$
$(d)$ falls down the rope nearly freely under gravity? (Ignore the mass of the rope).

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(A) Mass of the monkey,$m = 40 \, kg$.
Maximum tension that the rope can bear,$T_{\max} = 600 \, N$.
Case $(a)$: Acceleration of the monkey,$a = 6 \, m \, s^{-2}$ upward.
Using Newton's second law of motion,$T - mg = ma$.
$T = m(g + a) = 40(10 + 6) = 40 \times 16 = 640 \, N$.
Since $T > T_{\max}$,the rope will break in this case.
Case $(b)$: Acceleration of the monkey,$a = 4 \, m \, s^{-2}$ downward.
Using Newton's second law of motion,$mg - T = ma$.
$T = m(g - a) = 40(10 - 4) = 40 \times 6 = 240 \, N$.
Since $T < T_{\max}$,the rope will not break.
Case $(c)$: Uniform speed of $5 \, m \, s^{-1}$,so acceleration $a = 0$.
$T = mg = 40 \times 10 = 400 \, N$.
Since $T < T_{\max}$,the rope will not break.
Case $(d)$: Free fall,so $a = g$.
$T = m(g - g) = 0 \, N$.
Since $T < T_{\max}$,the rope will not break.

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