A3-2. Kinetic and Potential Energy
1. Kinetic Energy and the Work-Kinetic Energy Theorem
Kinetic energy ($K$ or $E_k$) is the energy an object possesses purely due to its motion. The magnitude of this energy depends on both the mass of the object and the square of its speed.
Kinetic Energy ($E_k$):
The Work-Kinetic Energy Theorem: The net work done on an object by all external forces equals its exact change in kinetic energy.
Derivation from Newton's Second Law & Kinematics:
Consider a constant net force $F$ acting on a mass $m$ over a displacement $S$.
From Newton's Second Law: $a = \dfrac{F}{m}$. Using the kinematic equation: ${v_f}^2 = {v_i}^2 + 2aS$ and substitute $a$
$$
{v_f}^2 = {v_i}^2 + 2\left(\dfrac{F}{m}\right)S
$$
Rearrange for $FS$, which is Work, to get
$$
2\left(\dfrac{F}{m}\right)S = {v_f}^2 - {v_i}^2 \implies FS = \dfrac{1}{2} m {v_f}^2 - \dfrac{1}{2} m {v_i}^2
$$
Therefore, $W_{net} = \Delta E_k$
Note: This holds true for variable forces as well, by breaking the displacement into infinitesimally small segments ($\Sigma F \Delta x$).
Example 1: The Falling Diver (Work-Kinetic Theorem)
Problem: A diver of mass $m$ steps off a platform of height $h$ from rest. (Assume $g$ is constant and air resistance is negligible).
- Use the Work-Kinetic Energy theorem to find the speed $v$ at which they hit the water.
- Calculate the work done by gravity specifically during the 2nd second of the fall (assuming they haven't hit the water yet), and the corresponding change in kinetic energy.
Solution:
- Part 1: Speed upon impact
The only force doing work is gravity ($W = mg \times h$).
$W_{net} = \Delta E_k \implies mgh = \dfrac{1}{2} m v^2 - 0 \implies v = \sqrt{2gh}$ - Part 2: Work and $\Delta E_k$ in the 2nd second
First, find the displacement during the 2nd second (from $t=1$ to $t=2$).
Displacement at $t=1$: $$ y_1 = \dfrac{1}{2}g(1)^2 = \dfrac{1}{2}g $$ Displacement at $t=2$: $$ y_2 = \dfrac{1}{2}g(2)^2 = 2g $$ Distance fallen in the 2nd second: $$ \Delta y = 2g - \dfrac{1}{2}g = \dfrac{3}{2}g $$ Work done by gravity: $$ W = F \times \Delta y = mg \times \left(\dfrac{3}{2}g\right) = \mathbf{\dfrac{3}{2}mg^2} $$ By the theorem, the change in kinetic energy $\Delta E_k$ during this specific second is exactly equal to the work done: $$ \mathbf{\Delta E_k = \dfrac{3}{2}mg^2} $$
2. Conservative Forces and Potential Energy
Potential energy can only be defined for conservative forces (like gravity and spring forces). For these forces, the work done in moving an object between two points is completely independent of the path taken. If you move an object in a closed loop, the total work done by a conservative force is zero.
Non-conservative forces (like friction or air resistance) drain mechanical energy from the system, converting it into heat. You cannot define a "frictional potential energy" because the work done depends on the path length.
3. Gravitational Potential Energy (Uniform vs. General)
A. Uniform Gravitational Field (Near Earth's Surface)
Close to the Earth's surface, the gravitational acceleration $g$ is effectively constant. We can choose any convenient height to be our reference plane ($h=0$).
The work done by gravity is equal to the negative change in potential energy:
$W_g = U_i - U_f = -\Delta U$
When an object falls, gravity does positive work, and the object's potential energy decreases.
B. General Gravitational Field HL ONLY
When moving far away from the Earth (e.g., satellites), gravity is a variable force that weakens according to Newton's Law of Universal Gravitation. We define the zero potential energy reference point to be at infinity.
Universal Gravitational Potential Energy ($U_g$):
- $G$: Universal Gravitational Constant ($6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$)
- $M$: Mass of the central body (e.g., Earth)
- $m$: Mass of the orbiting object
- $r$: Distance from the center of the planet
Why is it negative? Because gravity is an attractive force. If potential energy is zero at infinity, any object falling closer to Earth loses potential energy. Since it starts at zero and decreases, it must become negative. You must do positive external work to move an object from Earth's surface out to infinity.
Example 2: Universal Gravitational Potential Energy
Problem: A satellite of mass $m = 2000\text{ kg}$ orbits Earth. The mass of Earth is $M = 6.0 \times 10^{24}\text{ kg}$ and its radius is $R_E = 6.4 \times 10^6\text{ m}$. Using infinity as the zero-potential reference point, calculate the satellite's gravitational potential energy:
(1) When it is resting on Earth's surface.
(2) When it is in orbit at an altitude of $h = \dfrac{R_E}{2}$.
*(Use $G = 6.67 \times 10^{-11}\text{ N}\cdot\text{m}^2/\text{kg}^2$)*
Solution:
- Part 1: On the surface ($r = R_E$)
$$U_1 = -\dfrac{GMm}{R_E} = -\dfrac{(6.67 \times 10^{-11})(6.0 \times 10^{24})(2000)}{6.4 \times 10^6} \approx \mathbf{-1.25 \times 10^{11}\text{ J}}$$ - Part 2: In orbit ($r = R_E + \dfrac{R_E}{2} = 1.5 R_E$)
Since the distance from the center is $1.5$ times larger, the potential energy magnitude is divided by $1.5$.
$$U_2 = \dfrac{U_1}{1.5} = \dfrac{-1.25 \times 10^{11}}{1.5} \approx \mathbf{-8.33 \times 10^{10}\text{ J}}$$ - Notice that the potential energy increased (became a smaller negative number) as it moved away from Earth.
4. Elastic Potential Energy
Elastic potential energy is the energy stored within a material when it undergoes temporary deformation (stretching or compressing). The most common model is an ideal spring obeying Hooke's Law.
Elastic Potential Energy ($U_s$):
- $k$: Spring constant (or stiffness) in Newtons per metre (N/m). A higher $k$ means a stiffer spring that stores more energy for the same deformation.
- $x$: Extension or compression from the equilibrium point in metres (m).
Example 3: Energy Stored in a Spring
Problem: A mass of 14 g is suspended from a fixed spring. The spring has a spring constant of 3.0 N/m. Calculate the energy stored in the spring when it is extended by 18 mm.
Solution:
- Step 1: Convert units to standard SI base units.
Extension $x = 18\text{ mm} = 0.018\text{ m}$.
(Note: The mass of 14 g is distractor information since the exact equilibrium extension distance is directly provided.) - Step 2: Calculate Elastic Potential Energy.
$$U_s = \dfrac{1}{2} k x^2$$
$$U_s = \dfrac{1}{2} (3.0)(0.018)^2$$
$$U_s = 1.5 \times 0.000324 = \mathbf{4.86 \times 10^{-4}\text{ J}}$$