F=20i^+10j^
a=mF=220i^+10j^⇒10i^+5j^
∴s=21at2=21(10i^+5j^)×(10)2
⇒50(10i^+5j^)m
∴ Displacement along x-axis
⇒50×10⇒500m
JEE Main 2021 — Physics Mechanics
A boy pushes a box of mass 2kg with a force F=(20i^+10j^)N on a frictionless surface. If the box was initially at rest, then _______ m is displacement along the x-axis after 10s
Held on 26 Feb 2021 · Verified 6 Jul 2026.
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