Kinetic energy is the energy an object possesses due to its motion. It is one of the most important quantities in physics because of the work–energy theorem: the net work done on an object equals its change in kinetic energy. This simple yet profound relation connects force (dynamics) to energy.
Kinetic energy definition
The kinetic energy of a moving object of mass m with speed v is:
\[ KE = \frac{1}{2} m v^2 \]
where m is mass [kg], v is speed [m/s], and KE is kinetic energy [J].
Key observations:
- Kinetic energy is always positive (since \( v^2 \geq 0 \))
- It depends on the square of speed — doubling speed quadruples kinetic energy
- It is independent of direction — only the magnitude
v = |v|matters - At rest (
v = 0), kinetic energy is zero
Numerical example: fast-moving bullet
A 10 g bullet (0.01 kg) travels at 400 m/s.
\[ KE = \frac{1}{2} m v^2 = \frac{1}{2} \times 0.01 \times (400)^2 = 0.005 \times 160{,}000 = 800 \text{ J} \]
The bullet carries 800 J of kinetic energy. This is why bullets can penetrate material — the kinetic energy concentrates force over a small area.
The work–energy theorem
The work–energy theorem is the central result of this chapter. It states:
\[ W_{\text{net}} = \Delta KE = KE_f - KE_i = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]
where \( W_{\text{net}} \) is the net work (sum of all forces), \( KE_i \) is initial kinetic energy, and \( KE_f \) is final kinetic energy.
In words: The net work equals the change in kinetic energy.
Why this is true
The proof uses Newton's second law and calculus:
\[ F_{\text{net}} = ma = m\frac{dv}{dt} \]
Rearranging: \[ F_{\text{net}} \, dx = m \frac{dv}{dt} \, dx = m \frac{dv}{dx} \frac{dx}{dt} \, dx = m v \, dv \]
where we used \( v = dx/dt \).
Integrating both sides from initial position \( x_i \) to final position \( x_f \):
\[ \int_{x_i}^{x_f} F_{\text{net}} \, dx = \int_{v_i}^{v_f} m v \, dv = m \left[ \frac{v^2}{2} \right]_{v_i}^{v_f} = \frac{1}{2} m v_f^2 - \frac{1}{2} m v_i^2 \]
The left side is the net work, the right side is the change in kinetic energy. Thus:
\[ W_{\text{net}} = \Delta KE \]
Applications
Example 1: Accelerating a car
A 1500 kg car starts from rest (v_i = 0) and accelerates to 25 m/s due to the engine's thrust force. What is the net work done on the car?
\[ \Delta KE = \frac{1}{2} \times 1500 \times (25)^2 - \frac{1}{2} \times 1500 \times 0^2 \]
\[ = \frac{1}{2} \times 1500 \times 625 = 468{,}750 \text{ J} \approx 469 \text{ kJ} \]
The engine must do 469 kJ of net work. (In practice, more than this is needed because friction and air drag do negative work, so the engine's output power is higher.)
Example 2: Braking distance
A car traveling at 30 m/s brakes with a friction force of 6000 N. How far does it travel before stopping?
Method: Use the work–energy theorem. The friction force does negative work until the car stops.
\[ W_{\text{friction}} = \Delta KE = 0 - \frac{1}{2} \times 1500 \times (30)^2 \]
\[ = -\frac{1}{2} \times 1500 \times 900 = -675{,}000 \text{ J} \]
The friction force is 6000 N backward, so:
\[ W_{\text{friction}} = F_{\text{friction}} \times d = -6000 \times d \]
Setting these equal:
\[ -6000 \times d = -675{,}000 \]
\[ d = \frac{675{,}000}{6000} = 112.5 \text{ m} \]
The car travels 112.5 m before stopping.
Example 3: Spring collision
A 2 kg block slides into a spring on a frictionless surface. The block enters with speed 4 m/s and compresses the spring until it stops. If the spring constant is k = 200 N/m, how much does the spring compress?
Initial kinetic energy: \[ KE_i = \frac{1}{2} \times 2 \times (4)^2 = 16 \text{ J} \]
Work done by spring: From §7.2, when a spring compresses by distance \( x_0 \), the work it does is: \[ W_{\text{spring}} = -\frac{1}{2} k x_0^2 \]
Applying work–energy theorem (with final speed zero): \[ W_{\text{spring}} = KE_f - KE_i = 0 - 16 \]
\[ -\frac{1}{2} k x_0^2 = -16 \]
\[ \frac{1}{2} \times 200 \times x_0^2 = 16 \]
\[ 100 x_0^2 = 16 \]
\[ x_0^2 = 0.16 \]
\[ x_0 = 0.4 \text{ m} = 40 \text{ cm} \]
The spring compresses by 40 cm.
Kinetic energy and reference frames
Recall from Chapter 2 that velocity is relative to a reference frame. So is kinetic energy!
An object moving at 10 m/s in frame A and 0 m/s in frame B (moving alongside the object) has different kinetic energies in the two frames:
- Frame A: \( KE = \frac{1}{2} m (10)^2 = 50m \) J
- Frame B: \( KE = \frac{1}{2} m (0)^2 = 0 \) J
This is not a contradiction — it simply reflects that kinetic energy (like velocity) depends on the observer's frame. However, changes in kinetic energy are frame-independent if all observers use the same time interval.
Connection to §7.2 (Variable Forces)
In §7.2, we learned that work by a variable force requires integration: \( W = \int F_x \, dx \). The work–energy theorem tells us that this integral always equals the change in kinetic energy, regardless of how complicated the force function is. This is a powerful result: we don't always need to know the force as a function of position — if we know the initial and final speeds, we can find the work.
Connection to Chapter 5 (Newton's Laws)
Newton's second law relates force to acceleration: \( F = ma \). The work–energy theorem shows how this microscopic relation (force at each instant) accumulates into a macroscopic result (change in kinetic energy over a displacement). This is the first link between dynamics (forces) and energy.
What you should be able to do
After this section, you should be able to:
- Calculate the kinetic energy of an object given its mass and speed
- Apply the work–energy theorem to find net work from initial and final speeds
- Solve for unknown distances using the work–energy theorem when forces are given
- Understand that kinetic energy (like velocity) is frame-dependent
- Recognize that the work–energy theorem works for both constant and variable forces
Preview of §7.4: Potential Energy
The work–energy theorem connects net work to kinetic energy change. But some forces (like gravity and springs) are conservative: the work they do depends only on start and end positions, not the path.
For conservative forces, we can define a potential energy U(x) such that:
\[ W_{\text{conservative}} = -\Delta U = -(U_f - U_i) = U_i - U_f \]
This leads to the principle of mechanical energy conservation: \( E_{\text{mech}} = KE + U = \text{constant} \) (when only conservative forces act). This is §7.4.
📚 See also: Halliday Vol 1, Ch 7, §7.4 — Work and Kinetic Energy. 📖 Open reference: OpenStax University Physics Vol 1 — Chapter 7.2: Kinetic Energy and the Work-Energy Theorem. 🎓 Watch: MIT 8.01 Lecture 9 (Walter Lewin) — "Potential Energy, Kinetic Energy, and the Work-Energy Theorem".
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