krit.club logo

Work, Energy, and Power - Conservation of Mechanical Energy

Grade 11CBSEPhysics

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

🔑Concepts

•

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.

•

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.

•

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.

•

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.

•

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}.

•

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.