Physics - NLM Question with Solution | TestHub

Weight of the climber $mg=65\times 9.8=637\mathrm{N}$. Net downward force $=mg-T$, so the change in momentum equals the area between $mg$ and the $T$-$t$ graph.

From 0 to 0.2 s, $T$ falls from 637 N to 312 N, so $mg-T$ rises from 0 to 325 N. Impulse $=\frac{1}{2}(0.2)(325)=32.5\mathrm{N}\mathrm{s}$.

From 0.2 to 2 s, $T$ rises from 312 N to 637 N, so $mg-T$ falls from 325 N to 0. Impulse $=\frac{1}{2}(1.8)(325)=292.5\mathrm{N}\mathrm{s}$.

Total impulse $=325\mathrm{N}\mathrm{s}=mv\Rightarrow v=\frac{325}{65}=5m/s$.