After 10sec, 21mv2=81mv02
⇒v=2v0=5ms−1
Acceleration, a=−mkv2
⇒dtdv=−mkv2
⇒∫105v2dv=−mk∫010dt
−51+101=10−2−k×10
k=10−4kgm−1
JEE Main 2017 — Physics Mechanics
A body of mass m=10−2kg is moving in a medium and experiences a frictional force F=−kv2. Its initial speed is v0=10ms−1. After 10s its kinetic energy is 81mv02, then value of k will be:-
Held on 2 Apr 2017 · Verified 6 Jul 2026.
10−1kgm−1s−1
10−3kgm−1
10−3kgs−1
10−4kgm−1
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