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Work, Energy, and Power - Conservation of Mechanical Energy

Grade 11CBSEPhysics

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

🔑Concepts

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Total Mechanical Energy (EE) of a system is defined as the sum of its kinetic energy (KK) and potential energy (UU): E=K+UE = K + U.

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The Principle of Conservation of Mechanical Energy states that if only conservative forces (like gravity or spring force) perform work on a system, the total mechanical energy remains constant: K1+U1=K2+U2K_1 + U_1 = K_2 + U_2.

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A force is considered conservative if the work done by the force on a particle moving between two points is independent of the path taken. Examples include the gravitational force F⃗g\vec{F}_g and the electrostatic force F⃗e\vec{F}_e.

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Non-conservative forces, such as friction or air resistance, cause a change in the total mechanical energy of the system. The work done by non-conservative forces WncW_{nc} is equal to the change in mechanical energy: Wnc=ΔEW_{nc} = \Delta E.

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For a conservative force acting in one dimension, the force F(x)F(x) is related to the potential energy U(x)U(x) by the relation F(x)=−dUdxF(x) = -\frac{dU}{dx}.

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In a closed system with no external work or non-conservative forces, the change in kinetic energy is the negative of the change in potential energy: ΔK=−ΔU\Delta K = -\Delta U.

📐Formulae

E=K+UE = K + U

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

Ugrav=mghU_{grav} = mgh

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

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

F(x)=−dUdxF(x) = -\frac{dU}{dx}

ΔK+ΔU=Wnc\Delta K + \Delta U = W_{nc}

💡Examples

Problem 1:

A ball of mass m=0.5 kgm = 0.5\text{ kg} is dropped from the top of a building of height h=20 mh = 20\text{ m}. Using the law of conservation of energy, find its velocity vv just before it hits the ground. (Take g=10 m/s2g = 10\text{ m/s}^2)

Solution:

Initial energy at the top: Ei=Ki+UiE_i = K_i + U_i. Since it is dropped from rest, Ki=0K_i = 0. Thus, Ei=mghE_i = mgh. Final energy at the bottom: Ef=Kf+UfE_f = K_f + U_f. Taking the ground as reference level, Uf=0U_f = 0. Thus, Ef=12mv2E_f = \frac{1}{2}mv^2. By conservation of energy: mgh=12mv2mgh = \frac{1}{2}mv^2. Solving for vv: v=2gh=2×10×20=400=20 m/sv = \sqrt{2gh} = \sqrt{2 \times 10 \times 20} = \sqrt{400} = 20\text{ m/s}.

Explanation:

The gravitational potential energy at the maximum height is entirely converted into kinetic energy at the point of impact, assuming air resistance is negligible.

Problem 2:

A block of mass mm is pushed against a horizontal spring with spring constant kk, compressing it by a distance xx. When the block is released, it slides on a frictionless surface. What is the speed vv of the block when it leaves the spring?

Solution:

The initial energy stored in the compressed spring is the elastic potential energy: Ui=12kx2U_i = \frac{1}{2}kx^2. The initial kinetic energy Ki=0K_i = 0. When the spring returns to its natural length, the potential energy Uf=0U_f = 0 and the energy is transferred to the block as kinetic energy Kf=12mv2K_f = \frac{1}{2}mv^2. By conservation of energy: 12kx2=12mv2\frac{1}{2}kx^2 = \frac{1}{2}mv^2. This gives v=xkmv = x\sqrt{\frac{k}{m}}.

Explanation:

The elastic potential energy stored in the spring due to compression is converted into the kinetic energy of the block as the spring restores to its equilibrium position.

Conservation of Mechanical Energy - Revision Notes & Key Formulas | CBSE Class 11 Physics