Here are your friendly physics notes on springs, energy, and more – perfect for building a strong foundation! 😊“`html

Spring Force and Hooke’s Law

When you stretch or squeeze a spring, it pushes or pulls back! 🦾 This force is called the spring force (Fs). For an ideal spring:

  • Hooke’s Law says: Fs = -kx
  • k = spring constant (units: N/m). Stiff springs have large k; soft springs have small k.
  • x = displacement from equilibrium (stretch or squeeze distance).
  • The negative sign (−) means the spring force always opposes the displacement. 🔄

Potential Energy of a Spring

When you stretch/squeeze a spring, you store energy! ⚡ This is spring potential energy (V):

  • V(x) = \(\frac{1}{2}kx^2\)
  • At equilibrium (x = 0), V = 0.
  • Work done by the spring when stretched/squeezed to xm: \(W_s = -\frac{1}{2}kx_m^2\) (negative because the spring resists!).
  • Work done by you (external force) to stretch/squeeze it: \(W = +\frac{1}{2}kx_m^2\). 💪

Conservation of Mechanical Energy

In a spring-mass system (no friction!):

  • Total mechanical energy (E) is conserved: \(E = \frac{1}{2}kx^2 + \frac{1}{2}mv^2 = \text{constant}\) 🌟
  • At maximum stretch/squeeze (xm), speed v = 0 → all energy is potential: \(E = \frac{1}{2}kx_m^2\).
  • At equilibrium (x = 0), potential energy = 0 → all energy is kinetic → max speed: \(v_m = \sqrt{\frac{k}{m}} x_m\). 🚀

Example: Car hitting a spring

A car (mass = 1000 kg) moving at 18 km/h (= 5 m/s) hits a spring (k = 5.25 × 10³ N/m). Max compression of the spring?
Solution: Kinetic energy → spring potential energy!
\(K = \frac{1}{2}mv^2 = \frac{1}{2}(1000)(5)^2 = 12500 \text{ J}\)
At max compression: \(\frac{1}{2}kx_m^2 = 12500\) → \(x_m = \sqrt{\frac{2 \times 12500}{5250}} = 2 \text{ m}\).

When Friction is Present

Friction is a non-conservative force (energy isn’t fully stored/recovered). 😓 Use the work-energy theorem:

  • Work done by friction reduces mechanical energy.
  • Change in mechanical energy: \(\Delta E = E_f – E_i = W_{nc}\) (work by non-conservative forces).

Example: Same car with friction (μ = 0.5)

\(\frac{1}{2}mv^2 = \frac{1}{2}kx_m^2 + \mu mg x_m\)
Solve quadratic equation: \(kx_m^2 + 2\mu mg x_m – mv^2 = 0\) → \(x_m = 1.35 \text{ m}\) (less than frictionless case!).

Power: Rate of Doing Work

  • Average power: \(P_{av} = \frac{\text{Work done (W)}}{\text{Time taken (t)}}\)
  • Instantaneous power: \(P = \frac{dW}{dt} = \vec{F} \cdot \vec{v}\) 🔥
  • Units: watt (W) = 1 J/s. Also: 1 horsepower (hp) = 746 W.

Example: Elevator power

Elevator (1800 kg) moves up at 2 m/s. Friction force = 4000 N.
Motor force needed: \(F = mg + \text{friction} = (1800 \times 10) + 4000 = 22000 \text{ N}\)
Power: \(P = F \cdot v = 22000 \times 2 = 44000 \text{ W} = 59 \text{ hp}\).

NEET High-Yield Concepts 💡

  1. Hooke’s Law & Spring Energy: \(F_s = -kx\) and \(V = \frac{1}{2}kx^2\) are MUST-KNOWs!
  2. Energy Conservation: \(\frac{1}{2}kx_m^2 = \frac{1}{2}kx^2 + \frac{1}{2}mv^2\) (spring systems without friction).
  3. Work-Energy Theorem: With friction, \(W_{nc} = \Delta E\) (NEET loves springs + friction problems!).
  4. Power: \(P = F \cdot v\) (applied to elevators, vehicles, etc.).

Keep practicing energy problems – you’ve got this! 💪✨

“`These notes break down springs, energy conservation, friction, and power in a simple way – perfect for building confidence! Remember: Hooke’s Law and energy formulas are your best friends here. 😊