Physics - NLM Question with Solution | TestHub

During climbing, $T=900\mathrm{N}$ and $g=10\mathrm{m}{\mathrm{s}}^{-2}$ (as used in the printed solution).

$900-60(10)=60a_{\text{man }}$, so $a_{\text{man }}=5\mathrm{m}{\mathrm{s}}^{-2}$ upward.

$v_{\text{man }}(2)=5\times 2=10\mathrm{m}{\mathrm{s}}^{-1}$ upward.

$900-80(10)=80a_{\text{block }}$, so $a_{\text{block }}=1.25\mathrm{m}{\mathrm{s}}^{-2}$ upward.

$v_{\text{block }}(2)=1.25\times 2=2.5\mathrm{m}{\mathrm{s}}^{-1}$ upward.

Immediately after the man stops climbing, both are moving upward; the rope goes slack. Both then accelerate downward at g until the rope becomes taut again.

Therefore A, B and C are correct; D is false.