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JEE Advanced · Physics · 15. Oscillations

A particle of mass 1 kg is subjected to a force which depends on the position as F=-kxi^+yj^ kg m s-2 with k=1 kg s-2. At time t=0, the particle's position r=12i^+2j^ m and its velocity v=-2i^+2j^+2πk^ m s-1. Let vx and vy denote the x and the y components of the particle's velocity, respectively. Ignore gravity. When z=0.5 m, the value of xvy-yvx is ______ m2 s-1.

  1. A 1
  2. B 2
  3. C 3
  4. D 4
Verified Solution

Answer & Solution

Correct Answer

(C) 3

Step-by-step Solution

Detailed explanation

Given here: F=-kxi^+yj^ kg m s-2 and m=1 kg In x-direction, Fx=-x=max
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