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Forces and Motion - Hooke's Law

Grade 9IB

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

🔑Concepts

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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.

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The Spring Constant (kk) is a measure of the stiffness of the spring. A higher value of kk means the spring is stiffer and requires more force to stretch.

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The Extension (xx) is the change in length of the object, calculated as x=L−L0x = L - L_0, where LL is the new length and L0L_0 is the original length.

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The Limit of Proportionality is the point beyond which the linear relationship between force and extension no longer holds (the graph is no longer a straight line).

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Elastic Deformation: The object returns to its original shape and size once the load is removed.

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Plastic Deformation: The object is permanently stretched and does not return to its original shape even after the force is removed.

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On a Force-Extension graph, the gradient of the linear section represents the spring constant (kk).

📐Formulae

F=kxF = kx

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

x=L−L0x = L - L_0

Ee=12kx2E_e = \frac{1}{2}kx^2

💡Examples

Problem 1:

A spring has an original length of 12 cm12\text{ cm}. When a force of 6 N6\text{ N} is applied, the length of the spring increases to 15 cm15\text{ cm}. Calculate the spring constant (kk) in N/m\text{N/m}.

Solution:

x=15 cm−12 cm=3 cmx = 15\text{ cm} - 12\text{ cm} = 3\text{ cm} x=0.03 mx = 0.03\text{ m} k=Fxk = \frac{F}{x} k=6 N0.03 m=200 N/mk = \frac{6\text{ N}}{0.03\text{ m}} = 200\text{ N/m}

Explanation:

First, calculate the extension by subtracting the original length from the final length. Convert the extension from centimeters to meters to ensure the spring constant is in standard units (N/m\text{N/m}). Finally, divide the force by the extension.

Problem 2:

A spring with a spring constant of 50 N/m50\text{ N/m} is stretched by 0.2 m0.2\text{ m}. Calculate the elastic potential energy stored in the spring.

Solution:

Ee=12kx2E_e = \frac{1}{2}kx^2 Ee=0.5×50 N/m×(0.2 m)2E_e = 0.5 \times 50\text{ N/m} \times (0.2\text{ m})^2 Ee=25×0.04E_e = 25 \times 0.04 Ee=1 JE_e = 1\text{ J}

Explanation:

Use the formula for elastic potential energy. Substitute the given values for kk and xx. Remember to square the extension (xx) before multiplying by the other factors.