a=mF=22i^+3j^+5k^
=i^+1.5j^+2.5k^
τ=ut+21at2
=0+21(i^+1.5j^+2.5k^)(16)
=8i^+12j^+20k^
b=12
JEE Main 2021 — Physics Mechanics
A body of mass 2kg moves under a force of (2i^+3j^+5k^)N It starts from rest and was at the origin initially. After 4s, its new coordinates are (8,b,20). The value of b is ______.
(Round off to the Nearest Integer)
Held on 16 Mar 2021 · Verified 6 Jul 2026.
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