Elastic PE — Revision Notes

⚡ 30-Second Revision

Definition — Energy stored in a deformed elastic object.
Hooke's Law — $F_s = -kx$ (restoring force is proportional to displacement).
Spring Constant (k) — Stiffness of spring, units N/m.
Elastic Potential Energy (EPE) — $U_e = \frac{1}{2}kx^2$.
Units — Joules (J).
Always Positive — $U_e \ge 0$.
Energy Conservation — $K + U_g + U_e = \text{constant}$ (if no non-conservative forces).
Series Springs — $1/k_{eff} = 1/k_1 + 1/k_2 + \dots$
Parallel Springs — $k_{eff} = k_1 + k_2 + \dots$

2-Minute Revision

Elastic potential energy (EPE) is the energy an elastic object, like a spring, stores when it's stretched or compressed. This storage happens because an external force does work against the spring's internal restoring force, which tries to bring it back to its original shape.

The key principle is Hooke's Law, $F = kx$, where $F$ is the force, $x$ is the displacement from equilibrium, and $k$ is the spring constant (a measure of stiffness). The formula for EPE is $U_e = \frac{1}{2}kx^2$.

Notice the $x^2$ term, meaning doubling the displacement quadruples the stored energy. EPE is always positive, as work is always done to deform the spring. This energy is crucial for understanding energy transformations, often converting into kinetic energy (e.

g., a released spring) or gravitational potential energy. Remember to apply the conservation of mechanical energy, $K + U_g + U_e = \text{constant}$, in problems involving springs, masses, and heights, ensuring consistent units throughout your calculations.

5-Minute Revision

Elastic potential energy (EPE) is the energy stored in an elastic system, such as a spring, due to its deformation (stretching or compression). This energy is a direct consequence of the work done by an external force to overcome the internal restoring forces of the material.

The fundamental relationship governing this is Hooke's Law, which states that the restoring force ($F_s$) is directly proportional to the displacement ($x$) from the equilibrium position: $F_s = -kx$.

Here, $k$ is the spring constant, indicating the stiffness of the spring. The negative sign signifies that the restoring force always opposes the displacement.

The formula for elastic potential energy is derived by calculating the work done by the external force ($F_{ext} = kx$) as the spring is deformed from $x=0$ to a final displacement $x$. Since the force is variable, integration is used: $U_e = \int_0^x kx \, dx = \frac{1}{2}kx^2$.

This formula highlights that EPE is always positive and depends quadratically on the displacement. For example, if a spring with $k=100\,\text{N/m}$ is stretched by $0.2\,\text{m}$, the stored energy is $U_e = \frac{1}{2}(100)(0.

2)^2 = 2\,\text{J}$. If stretched by$0.4\,\text{m}$, the energy becomes$U_e = \frac{1}{2}(100)(0.4)^2 = 8\,\text{J}$, which is four times the original energy.

EPE is a critical component in the principle of conservation of mechanical energy. In systems where only conservative forces (like spring force and gravity) act, the total mechanical energy ($K + U_g + U_e$) remains constant.

This principle is vital for solving problems involving energy transformations, such as a mass falling onto a spring or a block oscillating on a spring. For instance, if a $1\,\text{kg}$ mass compresses a spring ($k=500\,\text{N/m}$) by $0.

1\,\text{m}$after falling from a height, the initial gravitational potential energy ($mgh$) plus the additional gravitational potential energy lost during compression ($mgx$) equals the final elastic potential energy ($\frac{1}{2}kx^2$).

Always ensure all quantities are in SI units for accurate calculations. Remember the effective spring constants for series ($1/k_{eff} = \sum 1/k_i$) and parallel ($k_{eff} = \sum k_i$) combinations.

Prelims Revision Notes

Elastic potential energy ($U_e$) is the energy stored in an elastic object due to its deformation. It's a scalar quantity measured in Joules (J). The fundamental relationship is Hooke's Law: $F_s = -kx$, where $F_s$ is the restoring force, $k$ is the spring constant (N/m), and $x$ is the displacement from equilibrium (m). The negative sign indicates the restoring force opposes displacement. This law is valid only within the elastic limit of the material.

The formula for elastic potential energy is $U_e = \frac{1}{2}kx^2$. This is derived from the work done by an external force ($F_{ext} = kx$) to deform the spring. Since the force is variable, work is calculated via integration: $W = \int_0^x kx \, dx = \frac{1}{2}kx^2$. This implies that $U_e$ is always positive, regardless of whether the spring is stretched or compressed, because $x$ is squared. Doubling the displacement quadruples the stored energy.

Key Applications & Concepts for NEET:

1
Direct Calculation — Given $k$ and $x$, calculate $U_e$. Or, given $F$ and $x$, first find $k = F/x$, then calculate $U_e$.
2
Energy Conservation — This is a very common problem type. The total mechanical energy ($E = K + U_g + U_e$) is conserved if only conservative forces (gravity, spring force) are acting. For example, a mass falling onto a spring: Initial $U_g$ converts to $U_e$ and additional $U_g$ as the spring compresses. Set a reference level (e.g., lowest compression point) for $U_g=0$.
3
Oscillations (SHM) — In a mass-spring system, energy continuously converts between $U_e$ and kinetic energy ($K = \frac{1}{2}mv^2$). At maximum displacement (amplitude $A$), $K=0$ and $E_{total} = \frac{1}{2}kA^2$. At equilibrium ($x=0$), $U_e=0$ and $E_{total} = \frac{1}{2}mv_{max}^2$. Thus, $\frac{1}{2}kA^2 = \frac{1}{2}mv_{max}^2$.
4
Spring Combinations

* Series: $1/k_{eff} = 1/k_1 + 1/k_2 + \dots$ (Force is same, total extension is sum). * Parallel: $k_{eff} = k_1 + k_2 + \dots$ (Displacement is same, total force is sum).

1
Graphical Interpretation — The area under the Force vs. Displacement graph ($F_{ext}$ vs. $x$) represents the stored elastic potential energy. This area is a triangle, leading to the $\frac{1}{2} \times \text{base} \times \text{height}$ formula.

Common Mistakes to Avoid:

Not converting units (e.g., cm to m).
Forgetting to square $x$ in $U_e = \frac{1}{2}kx^2$.
Incorrectly applying energy conservation, especially in vertical spring problems (remember the $mgx$ term for gravitational potential energy change during compression/extension).
Confusing series and parallel spring formulas.

Vyyuha Quick Recall

Every Potential Energy Stored Keeps Xtra X-factor.

Elastic Potential Energy = $\frac{1}{2}$ K (spring constant) $\times$ X (displacement) $\times$ X (displacement) = $\frac{1}{2}k x^2$