Question

Elastic potential energy in elastic body

• A. Elastic potential energy is a form of potential energy stored within an elastic body as a result of deformation.
• B. When an external force is applied to stretch, compress, bend, or twist an elastic material, work is done against the internal elastic restoring forces.
• C. For a system exhibiting linear elasticity, such as an ideal spring obeying Hooke's Law ($F = -kx$), the elastic potential energy ($U$) stored when the spring is displaced by a distance $x$ from its equilibrium position is given by: $U = \frac{1}{2}kx^2$.
• D. The energy is zero at equilibrium ($x=0$) and increases quadratically with displacement.

Answer 1

Answer: (A), (B), (C), (D)

Elastic potential energy is the energy stored in an elastic body due to its deformation. For a spring, this is given by the formula $U = \frac{1}{2}kx^2$, where $k$ is the spring constant and $x$ is the displacement from equilibrium.

Answer 2

Elastic potential energy is a form of potential energy stored within an elastic body as a result of deformation. When an external force is applied to stretch, compress, bend, or twist an elastic material, work is done against the internal elastic restoring forces. This work done is stored as elastic potential energy within the material. When the deforming force is removed, the body tends to return to its original shape, releasing this stored energy.

For a system exhibiting linear elasticity, such as an ideal spring obeying Hooke's Law ($F = -kx$, where $F$ is the restoring force, $k$ is the spring constant, and $x$ is the displacement from equilibrium), the elastic potential energy ($U$) stored when the spring is displaced by a distance $x$ from its equilibrium position is given by:

$U = \frac{1}{2}kx^2$

This formula represents the work done to deform the spring from its equilibrium position to a displacement $x$. The energy is zero at equilibrium ($x=0$) and increases quadratically with displacement.