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Laws of Motion question

2005 · Shift 0 · Q175
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Laws of Motion question

2005 · Shift 0 · Q175

JEE MainPhysicsLaws of MotionMCQ+4 / −1
A particle of mass 0.3 kg subjected to a force F=−kxF=-kxF=−kx with k=15N/mk=15N/mk=15N/m. What will be its initial acceleration if it is released from a point 20 cm away from the origin?
  1. A
    15    m/s215\,\,\,\,m/{s^2}15m/s2
  2. B
    3   m/s23\,\,\,m/{s^2}3m/s2
  3. C
    10   m/s210\,\,\,m/{s^2}10m/s2
  4. D
    5   m/s25\,\,\,m/{s^2}5m/s2
View written solutionFree

Correct answer: C

  1. Given data

    • Mass of particle: m=0.3 kgm = 0.3\,\text{kg}m=0.3kg
    • Force law: F=−kxF = -kxF=−kx
    • Spring constant: k=15 N/mk = 15\,\text{N/m}k=15N/m
    • Initial displacement from origin: x=20 cm=0.2 mx = 20\,\text{cm} = 0.2\,\text{m}x=20cm=0.2m

  2. Find the initial force Using Hooke’s law: F=−kx=−15×0.2=−3 NF = -kx = -15 \times 0.2 = -3\,\text{N}F=−kx=−15×0.2=−3N The negative sign shows the force is directed toward the origin.

  3. Find the initial acceleration By Newton’s second law: a=Fm=−30.3=−10 m/s2a = \frac{F}{m} = \frac{-3}{0.3} = -10\,\text{m/s}^2a=mF​=0.3−3​=−10m/s2

  4. Interpretation The acceleration is toward the origin, with magnitude: ∣a∣=10 m/s2|a| = 10\,\text{m/s}^2∣a∣=10m/s2

  5. Match with options The correct option is: C: 10 m/s2\boxed{\text{C: } 10\,\text{m/s}^2}C: 10m/s2​

  6. Comparison with stored answer Stored correct answer: C

    Our derived answer is also C, so they agree.

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