The area under the force displacement curve give the amount of work done. 
From work-energy theorem, W=ΔKE∴ At x=8 m, W= Area ABDO+ Area CEFD =20×5+10×3=130 J Using Eq. (i) ⇒130=21mv2=21×1000500v2⇒v=2130=22.8 ms−1≈23 ms−1 At x=12 m W= Area ABDO+ Area CEFD + Area FGHIJ + Area KLMJ W=20×5+10×3+(−20×2)+(21×−5×2)+10×2[∵ Area FGHIJ = Area FGIJ+ Area GHI]=100+30−40−5+20=105 J Using Eq. (i) ∴105=21×21×v2⇒v=2105≃20.6 ms−1
An object of mass 500 ~g, initially at rest acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the…
NEET UG 2019 — Physics Mechanics
2019mcqhard
An object of mass 500 g, initially at rest acted upon by a variable force whose X component varies with X in the manner shown. The velocities of the object a point X=8 m and X=12 m, would be the respective values of (nearly) 
Official previous-year question
Held on 30 Apr 2019 · Verified 9 Jul 2026.
Options
- A
18 m/s and 24.4 m/s
- B
23 m/s and 24.4 m/s
- C
23 m/s and 20.6 m/s
- D
18 m/s and 20.6 m/s
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