Work to Stretch a Spring

Work Needed to Stretch or Compress a Spring

Definitions

  • Natural Length: The length of a spring at rest.

Hooke's Law

  • Statement: The force (F) required to maintain a spring either stretched or compressed (x units beyond its natural length) is proportional to x.

    • Mathematically, this is represented as:[ F(x) = k \cdot x ]

      • Where k is the spring constant (greater than zero, also considered stiffness).

  • Restoring Force: The force exerted by the spring that is equal and opposite to the applied force:[ F(x) = -k \cdot x ]

Coordinate System Selection

  • Origin Selection: When analyzing spring problems, we must choose where to set the origin of our coordinate system.

    • Commonly, the origin is placed at the natural length of the spring.

    • Stretching is in the positive x direction, and compression is in the negative x direction.

  • Visualization: It is essential to draw a diagram of the situation to clarify the chosen coordinate system.

Calculating Work Done on a Spring

  • The work done (W) on the spring is calculated using the definite integral of the force function:[ W = \int_{a}^{b} F(x) , dx = \int_{a}^{b} k \cdot x , dx ]

    • Limits: A = initial position, B = final position.

Example Calculation

  1. Given Problem: A force of 40 pounds stretches a spring from 5 inches to 8 inches beyond its natural length.

  2. Units Conversion: Convert to feet as follows:

    • 5 inches = 5/12 feet

    • 8 inches = 8/12 feet = 2/3 feet

  3. Finding k: From the initial conditions, use 40 pounds to find k.[ 40 = k \cdot \frac{5}{12} \implies k = \frac{40 \cdot 12}{5} = 96 \text{ pounds/foot} ]

  4. Work Calculation: [ W = \int_{5/12}^{2/3} 96 \cdot x , dx ]

    • Resulting in:[ W = 96 \left( \frac{x^2}{2} \right) \bigg|_{5/12}^{2/3} ]

Work Done by Spring

  • When released: To find the work done by the spring when it returns to its natural length: [ W_{spring} = -\int_{b}^{a} k \cdot x , dx ]

    • Here, the upper bound (b) represents the starting stretch and the lower bound (a) is the natural length.

  • Example with known values from earlier calculation yields 64/3 foot-pounds.

Additional Example

  1. Work Required: 2 Joules to stretch a spring from 30 cm to 42 cm.

    • Convert, then solve using the known work formula and spring constant derived previously.

Average Force Calculation

  • Average force exerted during stretching can be calculated using:[ F_{avg} = \frac{1}{b-a} \int_{a}^{b} F(x) , dx ]

    • For example, after calculating work over an interval, you divide by the distance of the interval to find the average force.

Equilibrium Stretch with Applied Force

  • To find how far a force of 30 newtons would stretch the spring, use Hooke's law:[ F = k \cdot x \implies x = \frac{F}{k} ]

  • This requires the same spring constant, ultimately converting between meters and centimeters as necessary.