krit.club logo

Physics: Motion and Mechanics - Hooke's Law and Elasticity

Grade 8IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

•

Elasticity is the physical property of a material that allows it to return to its original shape and size after the deforming force (stress) is removed.

•

Plastic Deformation occurs when a material does not return to its original shape after the force is removed, resulting in a permanent change.

•

Hooke's Law states that the extension (xx) of an elastic object is directly proportional to the force (FF) applied to it, provided the limit of proportionality is not exceeded. This is mathematically written as F∝xF \propto x.

•

The Spring Constant (kk) is a measure of the stiffness of a spring. It is defined as the force required per unit of extension. Its unit is Newtons per meter (N/mN/m) or Newtons per centimeter (N/cmN/cm).

•

The Limit of Proportionality is the specific point beyond which the force and extension are no longer directly proportional (the graph is no longer a straight line).

•

The Elastic Limit is the point beyond which the object will no longer return to its original shape and will be permanently deformed.

•

Extension (xx) is calculated by subtracting the original length (L0L_0) from the new length (LL): x=L−L0x = L - L_0.

📐Formulae

F=k×xF = k \times x

k=Fxk = \frac{F}{x}

x=Lfinal−Loriginalx = L_{final} - L_{original}

W=m×gW = m \times g

💡Examples

Problem 1:

A spring has an original length of 12 cm12\text{ cm}. When a force of 10 N10\text{ N} is applied, the length of the spring becomes 15 cm15\text{ cm}. Calculate the spring constant (kk) of the spring.

Solution:

  1. Calculate the extension (xx): x=15 cm−12 cm=3 cmx = 15\text{ cm} - 12\text{ cm} = 3\text{ cm}
  2. Use Hooke's Law formula (F=kxF = kx): 10=k×310 = k \times 3
  3. Solve for kk: k=103≈3.33 N/cmk = \frac{10}{3} \approx 3.33\text{ N/cm}

Explanation:

First, we find how much the spring stretched by subtracting the original length from the final length. Then, we rearrange the Hooke's Law formula to solve for the spring constant.

Problem 2:

A spring with a spring constant k=50 N/mk = 50\text{ N/m} is compressed by 0.05 m0.05\text{ m}. What is the magnitude of the force applied to compress the spring?

Solution:

  1. Identify the values: k=50 N/mk = 50\text{ N/m} and x=0.05 mx = 0.05\text{ m}.
  2. Apply the formula: F=50×0.05F = 50 \times 0.05
  3. Calculate the result: F=2.5 NF = 2.5\text{ N}

Explanation:

By multiplying the spring constant by the distance of compression, we determine the force required to achieve that change in length.

Problem 3:

A mass of 200 g200\text{ g} is hung from a vertical spring. If g=10 N/kgg = 10\text{ N/kg} and the spring constant is 40 N/m40\text{ N/m}, calculate the extension of the spring in meters.

Solution:

  1. Convert mass to kg: 200 g=0.2 kg200\text{ g} = 0.2\text{ kg}.
  2. Calculate the weight (Force): F=0.2×10=2 NF = 0.2 \times 10 = 2\text{ N}
  3. Use x=Fkx = \frac{F}{k}: x=240x = \frac{2}{40}
  4. Final result: x=0.05 mx = 0.05\text{ m}

Explanation:

In vertical spring problems, the force is the weight of the object. We convert mass to kg to find weight in Newtons, then divide by the spring constant to find extension.