A force F = 0.12 N is applied to a spring and the spring elongates by 3 cm. The specific constant of the spring is?
Correct answer: D. 4 Nm-1
- A. 0.4 Nm-2
- B. 40 Nm-1
- C. 400 Nm-1
- D. 4 Nm-1
Explanation
When a spring is stretched/compressed by the application of a force, the relationship between the magnitude of the force applied and the elongation of the spring is given by Hooke's law: F=kx Where F is the magnitude of the spring applied k is the spring constant x is the elongation of the spring relative to its equilibrium position For the spring in this problem, we have: F = 0.12 N (force applied) x = 3 cm = 0.03 m (elongation of the spring) Therefore, we can solve the formula for k to find the spring constant: k=F/x k= 0.12/0.03 k= "4N/m"
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About Simple Harmonic Motion
Simple harmonic motion is oscillation in which acceleration is directly proportional to displacement and directed toward the equilibrium position. Work includes displacement, velocity, acceleration, phase, period, frequency, amplitude and energy, with applications to springs and simple pendulums. The restoring force and the conditions for SHM distinguish it from general periodic motion.
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