Once you draw a free-body diagram identifying all forces, Newton's second law F_net = ma directly yields the equations governing motion. Each coordinate direction yields one differential equation; solving these systematically gives acceleration, which you integrate to find velocity and position. This bridges the gap between force diagrams and kinematic equations.
Start with single-force cases, then progressively add forces (gravity + normal, then friction). Repeatedly practice: sketch diagram → identify axes → write ΣF_x = ma_x and ΣF_y = ma_y separately → solve algebraically.
You already know two things that together make this topic powerful. From free-body diagrams, you know how to systematically identify and represent every force acting on an object: weight, normal force, friction, tension, applied forces — each a vector with a specific direction. From Newton's second law, you know that the net force on an object equals its mass times its acceleration: F_net = ma. This topic is about bridging the two — taking the forces you have drawn and turning them into equations you can solve.
The key procedural insight is that vectors must be analyzed component by component. If you orient your coordinate axes wisely, you can decompose every force into its x- and y-components, then write Newton's second law separately for each axis: ΣF_x = ma_x and ΣF_y = ma_y. These are two independent equations. On an inclined plane, for example, gravity acts downward — but if you orient the x-axis along the slope, gravity's component along the slope is mg sin θ (driving the block down the slope) and the component perpendicular to the slope is mg cos θ (balanced by the normal force, giving N = mg cos θ). This decomposition is what makes inclined-plane problems tractable, and it illustrates a general principle: the choice of axes is yours, and a smart choice eliminates algebra.
The connection to calculus — which you have seen in your study of derivatives — is what turns "equations of motion" into a real mathematical object. Acceleration is the second derivative of position with respect to time: a = d²x/dt². So ΣF_x = ma_x is really the differential equation m(d²x/dt²) = ΣF_x. In the simplest case of constant forces, you can solve this by integration: integrating once gives velocity v(t) = v₀ + at, and integrating again gives position x(t) = x₀ + v₀t + ½at². These are the kinematic equations you may have encountered before — now you understand where they come from. They are the solutions to Newton's second law under constant force, not independent postulates.
The most important practical skill is setting up the problem correctly before solving anything. The choice of coordinate axes matters enormously: aligning one axis with the direction of acceleration (or with the surface of contact) often eliminates one equation from the problem. A common failure is working in a rotated coordinate system but forgetting to rotate all force components — especially dangerous for friction (which acts tangent to the contact surface) and for normal forces (which act perpendicular to it). The methodology is always: draw the diagram → choose axes → decompose every force → apply ΣF = ma per axis → solve algebraically. Each step is simple; the failure mode is skipping one.