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Work, Energy, and Power - The Potential Energy of a Spring

Grade 11CBSEPhysics

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

🔑Concepts

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A spring exerts a restoring force that is proportional to the displacement from its equilibrium position. This is known as Hooke's Law: Fs=−kxF_s = -kx, where kk is the spring constant.

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The spring constant kk is a measure of the stiffness of the spring. Its SI unit is N⋅m−1N \cdot m^{-1}.

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Elastic Potential Energy V(x)V(x) is the energy stored in a spring when it is deformed (stretched or compressed). It is defined as V(x)=12kx2V(x) = \frac{1}{2} kx^2.

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The work done by the spring force WsW_s during a displacement from xix_i to xfx_f is given by Ws=12kxi2−12kxf2W_s = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2.

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The spring force is a conservative force. The total mechanical energy (kinetic + potential) of a mass-spring system remains constant in the absence of friction: E=12mv2+12kx2=constantE = \frac{1}{2}mv^2 + \frac{1}{2}kx^2 = \text{constant}.

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At the equilibrium position (x=0x = 0), the potential energy is zero and the kinetic energy is at its maximum.

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At the maximum displacement (amplitude xmx_m), the kinetic energy is zero and the potential energy is at its maximum: E=12kxm2E = \frac{1}{2}kx_m^2.

📐Formulae

Fs=−kxF_s = -kx

V(x)=12kx2V(x) = \frac{1}{2}kx^2

Ws=−∫0xFsdx=∫0xkxdx=12kx2W_s = - \int_{0}^{x} F_s dx = \int_{0}^{x} kx dx = \frac{1}{2}kx^2

Wnet=12kxi2−12kxf2W_{net} = \frac{1}{2}kx_i^2 - \frac{1}{2}kx_f^2

k=∣F∣xk = \frac{|F|}{x}

💡Examples

Problem 1:

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

Solution:

Given: k=400 N/mk = 400 \, N/m and x=5 cm=0.05 mx = 5 \, cm = 0.05 \, m. Using the formula: V=12kx2V = \frac{1}{2}kx^2 V=12×400×(0.05)2V = \frac{1}{2} \times 400 \times (0.05)^2 V=200×0.0025V = 200 \times 0.0025 V=0.5 JV = 0.5 \, J

Explanation:

The potential energy is calculated by converting the displacement into SI units (meters) and applying the expression for elastic potential energy.

Problem 2:

A mass of 2 kg2 \, kg is attached to a spring with k=200 N/mk = 200 \, N/m. If the spring is stretched by 0.1 m0.1 \, m and released from rest, what is the maximum speed of the mass?

Solution:

By conservation of energy, Max Potential Energy = Max Kinetic Energy. 12kxm2=12mvmax2\frac{1}{2}kx_m^2 = \frac{1}{2}mv_{max}^2 200×(0.1)2=2×vmax2200 \times (0.1)^2 = 2 \times v_{max}^2 200×0.01=2×vmax2200 \times 0.01 = 2 \times v_{max}^2 2=2×vmax22 = 2 \times v_{max}^2 vmax2=1v_{max}^2 = 1 vmax=1 m/sv_{max} = 1 \, m/s

Explanation:

At the point of release, all energy is potential. At the equilibrium position, all energy is converted to kinetic energy, resulting in maximum velocity.