Acceleration can be written as, a=vdxdv.
Given: a=−2x(-ve sign is due to retardation)
Therefore,
vdxdv=−2x⇒vdv=(−2x)dx
Integrating both sides, we get
21(vf2−vi2)=−2(2x2)=x2⇒∣21m(vf2−vi2)∣=mx2
⇒∣ΔKE∣=0.01x2=(x10)−2
Hence, n=2.
JEE Main 2023 — Physics Mechanics
A particle of mass 10gmoves in a straight line with retardation 2x, where x is the displacement in SI units. Its loss of kinetic energy for above displacement is (x10)−nJ. The value of n will be ________.
Held on 6 Apr 2023 · Verified 6 Jul 2026.
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