AP Physics 1- 2.8 Spring Forces- Study Notes- New Syllabus
AP Physics 1-2.8 Spring Forces – Study Notes
AP Physics 1-2.8 Spring Forces – Study Notes -AP Physics 1 – per latest Syllabus.
Key Concepts:
- Force Exerted on an Object by an Ideal Spring
Force Exerted on an Object by an Ideal Spring
An ideal spring exerts a restoring force on an attached object that is directly proportional to the displacement of the object from the spring’s equilibrium (unstretched) position. This is described by Hooke’s Law.
Mathematical Expression (Hooke’s Law):
\( F_s = -k x \)
- \( F_s \): Restoring force exerted by the spring
- \( k \): Spring constant (N/m), measure of stiffness of the spring
- \( x \): Displacement of the spring from equilibrium (m)
- Negative sign (−) indicates the force is always opposite the direction of displacement.
Key Features:
- The force increases linearly with displacement (only valid in the elastic region).
- If the spring is compressed (\( x < 0 \)), the force pushes outward; if stretched (\( x > 0 \)), the force pulls inward.
- Restoring nature: always tends to bring the system back to equilibrium.
Potential Energy in a Spring:
\( U = \dfrac{1}{2} k x^2 \)
This represents the elastic potential energy stored in the spring when displaced by \( x \).
Example :
A spring has a constant \( k = 200 \, \text{N/m} \). If it is stretched by 0.05 m, find the restoring force and potential energy stored.
▶️Answer/Explanation
Force: \( F_s = -kx = -200(0.05) = -10 \, \text{N} \)
Potential Energy: \( U = \dfrac{1}{2} (200)(0.05^2) = 0.25 \, \text{J} \)
Answer: Restoring force = 10 N (toward equilibrium), Energy = 0.25 J
Example :
A 0.5 kg block is attached to a horizontal spring with spring constant \( k = 100 \, \text{N/m} \). The block is pulled 0.1 m from equilibrium and released. Find the maximum restoring force and the maximum potential energy stored in the spring.
▶️Answer/Explanation
Max Force: \( F_{max} = kx = 100(0.1) = 10 \, \text{N} \)
Max Potential Energy: \( U_{max} = \dfrac{1}{2} (100)(0.1^2) = 0.5 \, \text{J} \)
Answer: Max restoring force = 10 N, Max energy = 0.5 J
Example :
A vertical spring with constant \( k = 150 \, \text{N/m} \) is compressed by 0.08 m. Find the restoring force and elastic potential energy stored. If released, how much work will the spring do on the object?
▶️Answer/Explanation
Force: \( F_s = -kx = -150(0.08) = -12 \, \text{N} \)
Potential Energy: \( U = \dfrac{1}{2} (150)(0.08^2) = 0.48 \, \text{J} \)
Work done by spring (when released) = Energy stored = 0.48 J
Answer: Restoring force = 12 N (upward), Energy stored = 0.48 J, Work done = 0.48 J