In one line
Kinetics links motion to its cause: the net force on a particle equals its mass times its acceleration, \( \Sigma F = ma \), applied along each chosen direction.
Explained simply
Kinematics only describes motion (how fast, how far). Kinetics asks why: a heavier trolley needs a bigger push to speed up at the same rate. Newton's second law puts a number on that.
Technically, for a particle of mass m: \( \Sigma \vec{F} = m\vec{a} \). In rectangular components \( \Sigma F_x = ma_x \), \( \Sigma F_y = ma_y \); on a curved path \( \Sigma F_t = m\,dv/dt \), \( \Sigma F_n = mv^2/\rho \).
How it works
- Draw the FBD with all real forces: weight mg, normal reaction, friction, tension, applied forces.
- Choose axes: one along the direction of acceleration helps.
- Write \( \Sigma F = ma \) along that axis, and equilibrium (\( \Sigma F = 0 \)) perpendicular to it if there is no acceleration there.
- During sliding, kinetic friction \( F = \mu_k N \) opposes motion.
- Solve for a, then use kinematics for v and s.
Key formula
\( \Sigma F = ma \): F in N, m in kg, a in m/s\(^2\) (1 N = 1 kg·m/s\(^2\)). Weight W = mg with g = 9.81 m/s\(^2\).
Quick example
A 20 kg block on a floor (\( \mu_k = 0.2 \)) is pulled by a horizontal 100 N force. N = 196.2 N, friction = 39.24 N. \( a = (100 - 39.24)/20 = 3.04 \) m/s\(^2\). Starting from rest, after 3 s: \( v = 9.11 \) m/s.
Common mistakes
- Using weight in N where mass in kg is needed (or vice versa) in ma.
- Writing ma as a force on the FBD and also in the equation (counting it twice).
- Using static instead of kinetic friction once the body is sliding.
How MU asks it
Underlies all 5 to 8-mark Module 5 numericals: blocks on planes, connected bodies, lifts. The theory version (3 marks) is 'State Newton's laws and the equation of motion'. Show the FBD, the equation along the motion, and the kinematics step.