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JEE Main 2005 (AIEEE) · Inertia, First & Second Laws

Inertia, First & Second Laws question from JEE Main 2005 (AIEEE)

Q39JEE Main 2005 · AIEEEEasy

Inertia, First & Second Laws

A particle of mass 0.3 kg subjected to a force F=−kxF = -kx with k=15 N/mk = 15\text{ N/m}. What will be its initial acceleration if it is released from a point 20 cm away from the origin?

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Answer: (c)

  1. At release ∣F∣=kx=15×0.2=3 N|F| = kx = 15 \times 0.2 = 3\text{ N}
  2. a=Fm=30.3=10 m/s2a = \frac{F}{m} = \frac{3}{0.3} = 10\text{ m/s}^2
  3. So the answer is (c) 10 m/s210\text{ m/s}^2.
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