m=1kg
F=6t
∴a=mF=6t
dtdv=6t
∴∫0v0dv=∫016t⋅dt
∴v0=6×212=3ms−1
Δw=kf−ki=21×1×9−0=4.5J
JEE Main 2017 — Physics Mechanics
A time dependent force F=6t acts on a particle of mass 1kg. If the particle starts from the rest, the work done by the force during the first 1sec will be:
Held on 2 Apr 2017 · Verified 6 Jul 2026.
18J
4.5J
22J
9J
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