Interactive physics simulator
Elastic Potential Energy: PE = 1/2kx²
Investigate elastic deformation energy. Drag spring blocks to map restoring force integrals, pull bowstrings to launch projectiles against shatterable targets, and drop jumpers onto trampolines to solve dynamic deflection equations.
Elastic Potential Energy Lab
Adjust spring constant, mass, displacement, and drop height to analyze mechanical energy conservation and quadratic energy curves.
Live Telemetry
- Spring Constant (k)
- 200 N/m
- Displacement (x)
- 1.00 m
- Restoring Force (F_s)
- -200.0 N
- Elastic PE (PE_e)
- 100.0 J
- Projectile Mass (m)
- 0.20 kg
- Draw Distance (x)
- 0.50 m
- Band Stiffness (k)
- 400 N/m
- Stored Energy (PE_e)
- 50.0 J
- Launch Velocity (v)
- 0.0 m/s
- Kinetic Energy (KE)
- 0.0 J
- Jumper Mass (m)
- 60 kg
- Drop Height (h)
- 3.0 m
- Trampoline Stiffness (k)
- 5000 N/m
- Current Position (y)
- 3.00 m
- Max Deflection (x_max)
- 0.00 m
- Peak Impact Force
- 0.0 N
- Stored EPE
- 0.0 J
What is Elastic Potential Energy?
Elastic Potential Energy (PEe) is the energy stored in a mechanical system when a solid material is temporarily deformed (stretched, compressed, bent, or twisted) by an external force. This stored energy is directly related to the work done on the material to change its shape, and it is fully released when the restoring forces return the material to its original equilibrium state.
The mathematical equation calculating the elastic potential energy stored in an ideal Hookean spring is:
Where:
- k is the spring constant or stiffness coefficient of the material, measured in Newtons per meter (N/m)
- x is the displacement distance of the material from its equilibrium position (rest position), measured in meters (m)
Hooke's Law & Variable Forces
Unlike gravitational potential energy where the force is constant (F_g = mg), the force required to stretch or compress a spring changes continuously with displacement. According to Hooke's Law, the restoring force (Fs) is directly proportional to displacement:
The negative sign indicates that the restoring force acts in the opposite direction of the displacement, trying to pull the object back to the center equilibrium point. Because the force varies linearly from 0 to kx, the total work done to deform the spring is calculated by integrating the force over the distance:
Visually, this work corresponds to the triangular area under the Force vs Displacement graph. The base of this triangle is the displacement x, and the height is the restoring force magnitude F = kx. The area of the triangle is:
Conservation & Projectile Launch
When a stretched or compressed spring is released, the stored elastic potential energy converts entirely into kinetic energy (KE) in the absence of friction or dissipative heat losses.
For a projectile of mass m launched from a spring or slingshot of stiffness k stretched by distance x, the maximum velocity occurs at the equilibrium point (x = 0):
This relation explains why doubling the draw distance (2x) doubles the launch velocity (and quadruples the launch energy), whereas doubling the projectile mass (2m) reduces the launch velocity by a factor of √(2).
Solved Numerical Examples
A spring with a spring constant k = 250 N/m is compressed by a displacement x = 0.12 m from its equilibrium position. Calculate the stored elastic potential energy.
View Step-by-Step Solution
- Identify the given values: Spring constant k = 250 N/m, displacement x = 0.12 m.
- Recall the elastic potential energy formula: PEe = 1/2 · k · x².
- Square the displacement distance: x² = (0.12 m)² = 0.0144 m².
- Substitute the values into the equation: PEe = 0.5 · 250 N/m · 0.0144 m².
- Calculate the final energy: PEe = 1.8 Joules.
An archer pulls back a bow string by a distance of 0.6 m. The bow behaves like a spring with an effective force constant k = 300 N/m. A 0.05 kg arrow is launched from rest. Neglecting friction, calculate the maximum launch speed of the arrow.
View Step-by-Step Solution
- Identify the given values: Draw distance x = 0.6 m, spring constant k = 300 N/m, arrow mass m = 0.05 kg.
- Calculate the initial elastic potential energy stored in the bow: PEe = 1/2 · k · x² = 0.5 · 300 · (0.6)² = 54 Joules.
- Apply conservation of energy: The stored elastic potential energy is fully converted into the kinetic energy of the arrow: PEe = KE ⇒ 1/2 · k · x² = 1/2 · m · v².
- Simplify the energy equation to solve for velocity (v): v = x · √(k/m).
- Substitute values into the speed equation: v = 0.6 · √(300 / 0.05) = 0.6 · √(6000).
- Calculate final velocity: v = 0.6 · 77.46 m/s ≈ 46.48 m/s.
A professional diver of mass m = 70 kg drops from a height of h = 4.0 meters onto a trampoline with stiffness k = 8000 N/m. Solve for the maximum deflection (depression) x<sub>max</sub> of the trampoline bed. (Assume g = 9.8 m/s²)
View Step-by-Step Solution
- Identify the given values: Diver mass m = 70 kg, drop height h = 4.0 m, trampoline stiffness k = 8000 N/m, gravity g = 9.8 m/s².
- Set the reference datum (y = 0) at the undeformed trampoline bed. When the trampoline is depressed by a distance xmax, the diver has fallen a total vertical distance of h + xmax.
- Apply conservation of energy: The gravitational potential energy lost by the diver from the drop point to the lowest point is fully stored as elastic potential energy in the trampoline bed: mg(h + xmax) = 1/2 · k · xmax².
- Rearrange the energy balance into a standard quadratic equation: 1/2 · k · xmax² - mgxmax - mgh = 0.
- Substitute numerical values: 1/2(8000)xmax² - (70 · 9.8)xmax - (70 · 9.8 · 4.0) = 0 ⇒ 4000xmax² - 686xmax - 2744 = 0.
- Apply the quadratic formula xmax = [-b ± √(b² - 4ac)] / 2a, where a = 4000, b = -686, and c = -2744.
- Compute the discriminant: D = (-686)² - 4(4000)(-2744) = 470,596 + 43,904,000 = 44,374,596.
- Calculate positive deflection: xmax = [686 + √(44,374,596)] / 8000 = (686 + 6661.43) / 8000 ≈ 0.918 meters.
Conceptual Practice
A spring is stretched by 5 cm and stores 2 J of energy. If the stretch distance is increased to 15 cm, how much energy will be stored?
Show Explanation
Elastic potential energy scales quadratically with displacement (PEe ∝ x²). Since the displacement is tripled (from 5 cm to 15 cm), the stored energy increases by a factor of 3² = 9. Thus, the new stored energy is 2 J · 9 = 18 Joules.
Explain the negative sign in Hooke's Law (F_s = -kx) and why it does not make the elastic potential energy negative.
Show Explanation
The negative sign in Hooke's Law indicates that the restoring force is always opposite in direction to the displacement (e.g., if you stretch the spring to the right, it pulls back to the left). However, energy is work done against this force, and in the energy equation PEe = 1/2kx², the displacement is squared (x²), which is always positive, meaning stored potential energy is always positive.
A spring of stiffness k is cut exactly in half. What is the spring constant of each of the two halves, and how does this affect the energy stored when stretched by the same displacement x?
Show Explanation
Cutting a spring in half doubles its stiffness, so each half has a spring constant of 2k. This is because half the length requires twice the force to produce the same absolute displacement. Under the same displacement x, each half-spring will store twice as much energy (PEe = 1/2 · 2k · x² = kx²) compared to the original spring.
Why is a quadratic equation necessary to solve for the maximum trampoline depression in the drop lab rather than simply setting mgh = 1/2kx²?
Show Explanation
Setting mgh = 1/2kx² assumes that the loss in gravitational height is only the drop height h. However, as the trampoline bed deforms, the jumper continues to descend by an additional distance x. Therefore, the total vertical distance fallen is h + x. The conservation of energy must account for this extra gravitational work: mg(h + x) = 1/2kx², which leads directly to the quadratic equation 1/2kx² - mgx - mgh = 0.
Frequently Asked Questions
What is elastic potential energy?
Elastic potential energy is the stored energy in an object when it is temporarily deformed (stretched, compressed, bent, or twisted) by an external force. When the force is removed, the object returns to its original shape, releasing this energy.
What is the formula for elastic potential energy?
The formula is PEe = (1)/(2)kx2, where k is the spring constant in Newtons per meter (N/m), and x is the displacement or deformation distance in meters (m) from the spring's equilibrium position.
What is the spring constant k?
The spring constant (k) is a measure of the stiffness of a spring or elastic material. It represents the amount of force (in Newtons) required to stretch or compress the material by one meter. A higher k value indicates a stiffer spring.
What is Hooke's Law?
Hooke's Law states that the restoring force (Fs) exerted by a spring is directly proportional to its displacement (x) from its equilibrium position, acting in the opposite direction: Fs = -kx.
Why is displacement squared in the elastic potential energy formula?
Because the restoring force of a spring is variable (F = kx), the work done to compress/stretch it increases as it is deformed. The work is the area of the force-displacement triangle: Work = (1)/(2) · base · height = (1)/(2) · x · (kx) = (1)/(2)kx2.
What is the SI unit of elastic potential energy?
The SI unit of elastic potential energy is the Joule (J). One Joule is equivalent to one Newton-meter (1 N·m) or one kilogram meter squared per second squared (1 kg·m2/s2).
Does compressing a spring store more or less energy than stretching it by the same distance?
It stores the exact same amount of energy. In the formula PEe = (1)/(2)kx2, the displacement x is squared, so whether x is positive (stretched) or negative (compressed), x2 is always positive, resulting in identical energy storage.
What is the elastic limit of a material?
The elastic limit is the maximum amount of deformation a material can undergo without suffering permanent plastic deformation. If stretched beyond this limit, the spring will not return to its original shape and Hooke's law no longer applies.
How does elastic potential energy convert to kinetic energy during a launch?
In a frictionless system, releasing a stretched elastic band fully converts its stored potential energy into kinetic energy (1/2kx2 → 1/2mv2), accelerating the projectile to a maximum speed of v = x√(k/m) at the equilibrium point.
How does spring stiffness affect stored energy for a constant displacement?
Stored energy is directly proportional to spring stiffness (PEe ∝ k). For a constant displacement x, doubling the spring constant k doubles the stored potential energy.
How does displacement affect stored energy for a constant spring stiffness?
Stored energy scales quadratically with displacement (PEe ∝ x2). Doubling the stretch or compression distance (2x) quadruples the stored potential energy (22 = 4). Tripling the distance increases it by nine times (32 = 9).
Can elastic potential energy be negative?
No. Mass and spring constant k are always positive, and the displacement squared (x2) is always positive or zero. Therefore, PEe = (1)/(2)kx2 is always greater than or equal to zero.