A spring of force constant 400 N/m has an extension of 5 cm, the work done in extending it from 5 cm to 15 cm is:
Correct answer: B. 8 J
- A. 16 J
- B. 8 J
- C. 32 J
- D. 24 J
Explanation
Work done is found using the formula = ½ x k x [(x2)^2 - (x1)^2]= ½ x 400 x [(0.15)^2 - (0.05)^2]= 8 J
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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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