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Work, Energy, and Power - Kinetic and Potential Energy

Grade 11CBSEPhysics

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

🔑Concepts

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Kinetic Energy (KK) is the energy possessed by an object due to its motion. It is a scalar quantity and is always non-negative, defined as K=12mv2K = \frac{1}{2}mv^2.

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The Work-Energy Theorem states that the work done by the net force acting on a body is equal to the change in its kinetic energy: Wnet=ΔK=Kf−KiW_{net} = \Delta K = K_f - K_i.

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Potential Energy (UU) is the energy stored in a system by virtue of its position or configuration. It is defined only for conservative forces.

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Gravitational Potential Energy near the Earth's surface is given by U=mghU = mgh, where hh is the height above a chosen reference level.

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Elastic Potential Energy is stored in a deformed spring (either compressed or stretched) and is given by U=12kx2U = \frac{1}{2}kx^2, where kk is the spring constant.

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Conservative Forces are forces for which the work done depends only on the initial and final positions and not on the path taken (e.g., Gravitational force, Electrostatic force). For these forces, F=−dUdxF = -\frac{dU}{dx}.

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The Law of Conservation of Mechanical Energy states that if only conservative forces do work on a system, the total mechanical energy (E=K+UE = K + U) remains constant.

📐Formulae

K=12mv2K = \frac{1}{2}mv^2

Wnet=ΔK=12m(v2−u2)W_{net} = \Delta K = \frac{1}{2}m(v^2 - u^2)

Ugrav=mghU_{grav} = mgh

Uspring=12kx2U_{spring} = \frac{1}{2}kx^2

F(x)=−dUdxF(x) = -\frac{dU}{dx}

Ki+Ui=Kf+UfK_i + U_i = K_f + U_f

💡Examples

Problem 1:

A ball of mass 0.5 kg0.5\text{ kg} is dropped from a height of 20 m20\text{ m}. Calculate its velocity just before it hits the ground. (Take g=10 m/s2g = 10\text{ m/s}^2 and ignore air resistance).

Solution:

Using the Law of Conservation of Mechanical Energy: Einitial=EfinalE_{initial} = E_{final}. At the top, Ki=0K_i = 0 and Ui=mghU_i = mgh. At the bottom, Uf=0U_f = 0 and Kf=12mv2K_f = \frac{1}{2}mv^2. Therefore, mgh=12mv2  ⟹  v=2ghmgh = \frac{1}{2}mv^2 \implies v = \sqrt{2gh}. Substituting the values: v=2×10×20=400=20 m/sv = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\text{ m/s}.

Explanation:

The potential energy at the maximum height is entirely converted into kinetic energy at the ground level because only the conservative force of gravity is acting on the ball.

Problem 2:

A spring with spring constant k=500 N/mk = 500\text{ N/m} is compressed by 10 cm10\text{ cm}. Find the potential energy stored in the spring.

Solution:

Given k=500 N/mk = 500\text{ N/m} and x=10 cm=0.1 mx = 10\text{ cm} = 0.1\text{ m}. The elastic potential energy is U=12kx2U = \frac{1}{2}kx^2. Substituting the values: U=12×500×(0.1)2=250×0.01=2.5 JU = \frac{1}{2} \times 500 \times (0.1)^2 = 250 \times 0.01 = 2.5\text{ J}.

Explanation:

The work done in compressing the spring against the restoring force is stored as elastic potential energy. Note that the displacement must be converted to S.I. units (meters) before calculation.

Kinetic and Potential Energy - Revision Notes & Key Formulas | CBSE Class 11 Physics