4.5 Energy, work and power

a·b=a₁b₁+a₂b₂=|a||b|cosθ. It is zero for perpendicular vectors and gives the component of one vector along another after division by the reference magnitude.

Check both vectors are non-zero, use the principal angle and preserve units when a physical projection is requested.

For a=(2,1) and b=(1,2), a·b=4 and cosθ=4/5.

The dot product is not |a||b| without the cosine factor and is not a vector direction.

Near Earth’s surface, gravitational potential energy change is ΔE=mgh relative to a chosen reference level. Only differences matter, so the zero level can be selected for convenience.

Use energy conservation when no non-conservative work acts, and include kinetic and elastic terms as needed. Far from Earth, the uniform-field approximation may fail.

Raising 2 kg by 3 m where g=9.8 gives an increase of 58.8 J.

Potential energy itself is reference-dependent; a negative value is not automatically an error.

Forces arise from interactions such as weight, contact, tension, friction or thrust. A complete free-body diagram lets ΣF=ma be applied without double-counting.

Name the body, draw force directions and distinguish applied forces from resultant shorthand. Resolve only after the physical forces are identified.

For a block pulled by a rope on a rough plane, include weight, normal reaction, tension and friction before resolving along the plane.

Centripetal force is not an extra force; it is the name for the resultant inward force in circular motion.

Write z=x+iy and translate modulus or argument conditions into equations or inequalities in x and y. The resulting curve should match the geometric interpretation.

For |z−a|=r, expand to a circle; for equal distances, subtract squared distances to obtain a line. Check restrictions introduced by squaring or arguments.

|z−(1+i)|=|z−(−1+i)| simplifies to x=0, the vertical bisector of the two centres.

Squaring distances is safe for non-negative moduli, but argument equations still need branch and quadrant checks.

Acceleration is dv/dt, the instantaneous rate of change of velocity. Since velocity is ds/dt, acceleration is d²s/dt².

Keep signs and units consistent, and distinguish average acceleration Δv/Δt from the derivative at one instant.

If v(t)=t²−4t, then a(t)=2t−4; at t=3 the acceleration is 2 m s⁻².

Zero velocity does not imply zero acceleration: an object can be momentarily at rest while its velocity changes.