5.6 قوة الزنبرك

Understanding Hooke's Law and Spring Force

Introduction to Spring Mechanics

• The lesson begins with an introduction to the fifth unit on kinetic energy, work, power, and specifically focuses on spring force.

• A spring in equilibrium is described as having no change in length; external forces can stretch or compress it.

Characteristics of Spring Force

• The spring force acts towards the equilibrium position, opposing any external force applied to stretch or compress the spring.

• Hooke's Law states that the elastic (spring) force is directly proportional to the displacement from its equilibrium position but acts in the opposite direction.

Mathematical Representation

• The relationship between elastic force and displacement is expressed mathematically as F = -k Delta x , where k is a constant specific to each spring.

• The negative sign indicates that the spring force always opposes displacement. Units for k are Newton per meter (N/m).

Work Done by Spring Force

• Work done by a spring can be calculated using integration of force over distance: W = -1/2 k x^2 .

• If initial displacement ( x_initial ) is zero, then all potential energy stored in a compressed or stretched spring can be represented graphically.

Graphical Analysis of Forces

• A graph plotting external force against displacement shows a linear relationship until the limit of elasticity is reached; beyond this point, Hooke's law does not apply.

• The area under the curve represents work done on the spring, which translates into potential energy stored within it.

Energy Considerations

• When analyzing forces acting on a spring, if compression occurs without external work being applied, it requires an external effort to either extend or compress further.

• Negative work indicates that energy must be supplied externally for changes in length; thus potential energy stored in springs can be expressed as U = 1/2 k x^2 .

Applying Hooke's Law: Example Problems

Problem 1: Finding Spring Constant

• An example problem involves calculating the spring constant given an initial length and mass attached causing extension due to gravity.

Problem 2: Calculating Required Force

• Another scenario presents finding necessary forces when maintaining weight at specific positions above equilibrium while considering gravitational effects.

Problem 3: Compression Scenario

• A final example discusses how compression affects a massless spring placed horizontally under load and calculates displacements based on provided data.

What is the Speed of the Ball?

Understanding Work and Energy in Springs

• The discussion begins with calculating the speed of a ball, emphasizing that work done (W) relates to kinetic energy (KE) through the equation W = 1/2 k x^2 , where k is the spring constant and x is displacement.

• The initial velocity term becomes zero since the ball starts from rest. The calculation focuses on determining k using Hooke's Law, where force equals spring constant times displacement ( F = kx ).

• The spring constant is found to be 1.46 × 10³ N/m. Various work equations are discussed, including those involving force and displacement, highlighting their importance in solving problems related to springs.

• By substituting known values into the equations, it’s shown that after simplification, one can find final velocity ( v_f ) as 6.07 m/s for a mass of 75 kg.

Example Problems Involving Spring Compression

• A new example illustrates compressing a spring by distance h . The work done on this compression is expressed as W_h = 1/2 k h^2 .

• When doubling the compression distance to 2h , it’s noted that work increases by four times due to its quadratic relationship with displacement.

• Students are encouraged to use proportional reasoning for understanding how changes in distance affect work done on springs.

Calculating Spring Extension

• In another example, given a spring constant of 440 N/m and required work of 25 J, students learn how to calculate extension using direct substitution into the formula for work done on springs.

• Solving yields an extension of approximately 0.337 m when applying square root calculations based on rearranged formulas.

Comparing Work Done During Compression vs. Extension

• A scenario compares work during compression versus extension under ideal conditions without energy loss; both scenarios yield equal amounts of work due to identical parameters involved in each case.

Final Example: Energy Transformation from Potential to Kinetic

• An example discusses releasing a ball from a compressed spring; here, potential energy converts into kinetic energy upon release.

• Using conservation principles leads to finding that initial potential energy equals final kinetic energy resulting in a calculated speed of approximately 17.55 m/s after simplifications and substitutions are made throughout the process.