Physics · Ch 5 — Work, Energy and Power

Kinetic Energy and the Work-Energy Theorem

5.4

Kinetic Energy and the Work-Energy Theorem

Kinetic energy is defined as the energy a body possesses purely because of its motion:

K = \frac{1}{2}mv^2

where $m$ is the body's mass and $v$ is its speed. Kinetic energy is always a positive quantity (or zero, for a body at rest), since it depends on $v^2$ rather than on the velocity vector itself, and it shares the joule as its SI unit, exactly like work -- a fact that is no accident, and is made precise by the work-energy theorem below.

Deriving the work-energy theorem for a variable force. Consider a body of constant mass $m$ moving along a straight line under a net force $F(x)$ that may vary with position. Starting from the definition of work as an integral (§5.3) and using Newton's second law, $F = m\,dv/dt$ :

W_{\text{net}} = \int_{x_1}^{x_2} F\,dx = \int_{x_1}^{x_2} m\frac{dv}{dt}\,dx

The key step is to rewrite $dx$ in terms of $dv$ using the chain rule, $\dfrac{dv}{dt} = \dfrac{dv}{dx}\cdot\dfrac{dx}{dt} = v\dfrac{dv}{dx}$ , so that

W_{\text{net}} = \int_{x_1}^{x_2} m\left(v\frac{dv}{dx}\right)dx = \int_{v_1}^{v_2} mv\,dv = \left[\frac{1}{2}mv^2\right]_{v_1}^{v_2} = \frac{1}{2}mv_2^2 - \frac{1}{2}mv_1^2

The right-hand side is exactly $K_f - K_i$ , the final kinetic energy minus the initial kinetic energy. This gives the work-energy theorem:

W_{\text{net}} = \Delta K = K_f - K_i

The net (total) work done on a body by every force acting on it, added together as a single quantity, is always equal to the change produced in its kinetic energy -- exactly, whatever the individual forces are, and whether they are constant or variable, provided they are all correctly included in $W_{\text{net}}$ . …

Start your 14-day free trial to unlock the full solution →

← Previous · 5.3Work Done by a Variable Force Next · 5.5 →Power