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Work and Energy - Potential Energy of a Spring-advanced

Grade 9CBSE

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

🔑Concepts

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Elastic Potential Energy: This is the energy stored in an elastic object (like a spring) when it is deformed by an external force. For a spring, this energy is equal to the work done in compressing or stretching it.

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Hooke's Law: For an ideal spring, the restoring force FF is directly proportional to the displacement xx from the equilibrium position, expressed as F=−kxF = -kx, where kk is the spring constant.

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Spring Constant (kk): It represents the stiffness of the spring. Its SI unit is N/mN/m (Newton per meter). A higher value of kk indicates a stiffer spring.

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Work Done on a Spring: The work done to stretch or compress a spring by a distance xx is not simply F×xF \times x because the force changes as xx changes. It is calculated as the area under the Force-Displacement graph, which results in W=12kx2W = \frac{1}{2}kx^2.

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Restoring Force: When a spring is displaced, an internal force acts in the opposite direction to bring it back to its original shape. This is called the restoring force.

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Energy Transformation: In a spring-mass system, potential energy is maximum at extreme positions (maximum compression/extension) and kinetic energy is maximum at the equilibrium position (x=0x = 0).

📐Formulae

F=−kxF = -kx

U=12kx2U = \frac{1}{2}kx^2

W=12k(x22−x12)W = \frac{1}{2}k(x_2^2 - x_1^2)

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

💡Examples

Problem 1:

A spring with a spring constant of k=400 N/mk = 400\text{ N/m} is compressed by 5 cm5\text{ cm}. Calculate the elastic potential energy stored in the spring.

Solution:

Given: Spring constant k=400 N/mk = 400\text{ N/m} Compression x=5 cm=5100 m=0.05 mx = 5\text{ cm} = \frac{5}{100}\text{ m} = 0.05\text{ m} Using the formula for Potential Energy: U=12kx2U = \frac{1}{2}kx^2 U=12×400×(0.05)2U = \frac{1}{2} \times 400 \times (0.05)^2 U=200×0.0025U = 200 \times 0.0025 U=0.5 JU = 0.5\text{ J}

Explanation:

First, convert the displacement into SI units (meters). Then, substitute the values into the elastic potential energy formula. The result represents the work done in compressing the spring, now stored as energy.

Problem 2:

How much work is required to stretch a spring from an initial extension of 2 cm2\text{ cm} to a final extension of 4 cm4\text{ cm}, if the spring constant is 1000 N/m1000\text{ N/m}?

Solution:

Given: k=1000 N/mk = 1000\text{ N/m} x1=2 cm=0.02 mx_1 = 2\text{ cm} = 0.02\text{ m} x2=4 cm=0.04 mx_2 = 4\text{ cm} = 0.04\text{ m} Work done W=ΔU=12k(x22−x12)W = \Delta U = \frac{1}{2}k(x_2^2 - x_1^2) W=12×1000×((0.04)2−(0.02)2)W = \frac{1}{2} \times 1000 \times ((0.04)^2 - (0.02)^2) W=500×(0.0016−0.0004)W = 500 \times (0.0016 - 0.0004) W=500×0.0012W = 500 \times 0.0012 W=0.6 JW = 0.6\text{ J}

Explanation:

When a spring is already stretched, the work needed to stretch it further is the difference between the final potential energy and the initial potential energy.

Problem 3:

A mass attached to a spring stores 2 J2\text{ J} of energy when stretched by 0.1 m0.1\text{ m}. What is the spring constant kk?

Solution:

Given: U=2 JU = 2\text{ J} x=0.1 mx = 0.1\text{ m} Using U=12kx2U = \frac{1}{2}kx^2 2=12×k×(0.1)22 = \frac{1}{2} \times k \times (0.1)^2 2=12×k×0.012 = \frac{1}{2} \times k \times 0.01 4=k×0.014 = k \times 0.01 k=40.01k = \frac{4}{0.01} k=400 N/mk = 400\text{ N/m}

Explanation:

By rearranging the potential energy formula, we can solve for the spring constant kk if the energy and displacement are known.