Work by a Variable Force question from NEET 2019

Answer: (d)

1. Work = area under the $F$-$x$ graph, and $W = \Delta K$ (starts from rest, $m = 0.5\text{ kg}$).
2. $0 \to 8$ m: $W = 20 \times 5 + 10 \times 3 = 130\text{ J}$, so $v_8 = \sqrt{\frac{2 \times 130}{0.5}} = \sqrt{520} \approx 23\text{ m/s}$.
3. $8 \to 12$ m: the large negative dip (down to $-25$ N) outweighs the $+10$ N strip, so net work is negative (about $-24$ J) and the speed must drop below 23 m/s.
4. $K_{12} \approx 130 - 24 = 106\text{ J} \Rightarrow v_{12} = \sqrt{\frac{2 \times 106}{0.5}} \approx 20.6\text{ m/s}$
5. So the answer is (d) 23 m/s and 20.6 m/s.