{title2=$W_{\mathrm{net}}=\Delta T$}

For a <point mass> of constant <mass> $m$ observed in an <inertial frame>, the net <work> along its actual path equals its change in <kinetic energy>:
$$
W_{\mathrm{net}}=\int_{t_a}^{t_b}\mathbf F_{\mathrm{net}}(t)\cdot\mathbf v(t)\,dt
=\int_a^b\mathbf F_{\mathrm{net}}\cdot d\mathbf r
=\frac12m|\mathbf v_b|^2-\frac12m|\mathbf v_a|^2.
$$
The middle expression is a <line integral>; the <dot product> selects the component of <force> along the <displacement>. Positive net <work> increases <speed>, negative net <work> decreases it, and zero net <work> leaves the endpoint <speeds> equal. Equal endpoint <speeds> do not imply equal <velocities>: a perpendicular <force> can continually change direction while doing zero <work>. https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter13.pdf[MIT mechanics, Chapter 13, section 13.11]{ref}

The bridge to <Newton's second law> is short: take the <dot product> of $m\dot{\mathbf v}=\mathbf F_{\mathrm{net}}$ with <velocity>. The <chain rule> gives $d(m|\mathbf v|^2/2)/dt=\mathbf F_{\mathrm{net}}\cdot\mathbf v$, equating the rate of change of <kinetic energy> to instantaneous <power>. Integrating gives the theorem. No <conservative force> assumption entered this argument; <friction> also contributes to the net <work>.

For the <harmonic oscillator> with <spring> stiffness $k=2\,\mathrm{N/m}$, release a <mass> $m=0.5\,\mathrm{kg}$ from rest at $x=0.3\,\mathrm m$. The <spring> exerts $F=-kx$. Its <work> before the first passage through $x=0$ is
$$
W=\int_{0.3\,\mathrm m}^{0}(-kx)\,dx
=\frac12k(0.3\,\mathrm m)^2=0.09\,\mathrm J.
$$
Thus $mv^2/2=0.09\,\mathrm J$ gives <speed> $0.6\,\mathrm{m/s}$ there. The <velocity> is negative on this passage because the <position> is decreasing. The <energy> calculation alone determines its magnitude, so the direction still requires information about the motion.

For a time-independent <potential energy> $V$, the <conservative force> contributes <work> $-\Delta V$. Separating all other <work> as $W_{\mathrm{other}}$ gives $\Delta(T+V)=W_{\mathrm{other}}$. Consequently <conservation of mechanical energy> requires zero net other <work>; the <work-energy theorem> itself does not require that conservation. With <damping>, <mechanical energy> decreases even though the theorem remains valid. https://ocw.mit.edu/courses/8-01sc-classical-mechanics-fall-2016/mit8_01scs22_chapter14.pdf[MIT mechanics, Chapter 14: mechanical energy and nonconservative work]{ref}


 Work-energy theorem (source code)