Lecture on Hooke's Law and Spring Constant

Introduction

• Discussion on a problem related to Hooke's Law.
• Example: A spring stretches by 20 cm when a 150 N object is loaded onto it.

Key Concepts

Hooke's Law

• Formula: F = -KX
  • F: Force applied to the spring
  • K: Spring constant
  • X: Displacement or elongation of the spring
• The negative sign indicates that the restoring force by the spring opposes the applied force.

Understanding the Components

• Displacement (X): Measured from the spring's natural length (without mass) to its extended length (with mass).
• Spring Constant (K): A constant value for a given spring, indicating its stiffness.

Example Problem

Given

• Force = 150 N
• Elongation = 20 cm (convert to meters: 0.2 m)

Calculation of Spring Constant (K)

• Use the formula: K = F / X
• Calculation:
  • F = 150 N
  • X = 0.2 m
• Result: K = 150 N / 0.2 m = 750 N/m

Second Problem: New Mass

Given

• New mass = 20 kg
• Calculating new force:
  • Weight (F) = mass (m) x gravity (g)
  • F = 20 kg x 9.8 m/s² = 196 N

Finding New Displacement (X)

• Use Hooke's Law: X = F / K
• F = 196 N, K = 750 N/m
• Result: X = 196 N / 750 N/m = 0.26 m or 26 cm

Summary

• Hooke's Law helps calculate displacement or determine the spring constant.
• The spring constant remains unchanged unless the spring itself changes.

Conclusion

• Understanding the application of Hooke's Law for solving problems related to spring elongation and spring constant.
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