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Concepts Of Physics MCQ Edition [Volume 1]
Exam
Concepts Of Physics MCQ Edition [Volume 1]
Subject
Physics
Chapter
Newton's Laws of Motion

Concepts Of Physics MCQ Edition [Volume 1] · Physics · Newton's Laws of Motion

A force F = -kx , where k = 15 N/m , acts on a particle of mass 0.3...

A force F=−kxF = -kx, where k=15 N/mk = 15 \text{ N/m}, acts on a particle of mass 0.3 kg0.3 \text{ kg}. Determine the initial acceleration of the particle when it is released from the position x=20 cmx = 20 \text{ cm}.

Options

  1. A50 m/s250 \text{ m/s}^2
  2. B0.9 m/s20.9 \text{ m/s}^2
  3. C10 m/s210 \text{ m/s}^2
  4. D1000 m/s21000 \text{ m/s}^2

Correct answer

C. 10 m/s210 \text{ m/s}^2

Step-by-step solution

Given k=15 N/mk = 15 \text{ N/m}, m=0.3 kgm = 0.3 \text{ kg}, and x=20 cm=0.2 mx = 20 \text{ cm} = 0.2 \text{ m}.

The magnitude of the restoring force acting on the particle is given by ∣F∣=kx|F| = kx.

Substituting the given values:

∣F∣=15×0.2=3 N|F| = 15 \times 0.2 = 3 \text{ N}

The initial acceleration of the particle is a=∣F∣ma = \dfrac{|F|}{m}.

a=30.3=10 m/s2a = \dfrac{3}{0.3} = 10 \text{ m/s}^2
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