In the late 1600s and early 1700s, Isaac Newton published the Principia Mathematica<\/em> in which he built upon discoveries of scientists (at the time called \u201cnatural philosophers\u201d) before him, laying the foundation of classical mechanics. In the Principia<\/em>, he described three laws of motion:<\/p>\n\n\n\n We have already discussed the first law in the context of energy; if you understand inertia, you can explain many phenomena you observe. The second law is a mathematical statement which we can use to quantitatively solve problems. The third law, as we\u2019ll see later, is a result of conservation of momentum.<\/p>\n\n\n\n Let’s consider situations where an object’s mass remains constant (which is often the case), and apply Newton’s second law:<\/p>\n\n\n \\[ Since the mass is not changing, we can factor it out:<\/p>\n\n\n \\[ The rate of change of velocity is acceleration:<\/p>\n\n\n \\[ This is a specific case of Newton’s second law; it only applies when the mass of a system is not changing.<\/p>\n\n\n\n Remember that force, momentum and acceleration are all vector<\/em> quantities. Breaking them up into components gives us:<\/p>\n\n\n \\[ The unit of force is named the newton<\/em> (N) in reference to Isaac Newton. Using Newton’s second law, we can break the newton down into base SI units: N = kg\u00b7m\/s2<\/sup>. One newton is approximately the force required to push down one key on a computer keyboard.<\/p>\n\n\n\n A 1380 kg car is moving due East with an initial speed of 27 m\/s. After 8 s, the car has slowed down to 17 m\/s. Find the net force acting on the car.<\/p>\n\n\n\n Let’s call due East the \\(+x\\) direction. There are no \\(y\\) components of the car’s motion. Since this is a one-dimensional problem, we simply use + and – signs to indicate direction instead of using vector notation.<\/p>\n\n\n \\[ The negative value for the net force indicates that it is applied in the opposite direction as the car’s motion.<\/p>\n\n\n\n
\n\\begin{align}
\n \\vec{F}_\\textit{net} &= \\frac{\\Delta\\vec{p}}{\\Delta t} \\\\
\n &= \\frac{\\vec{p}_f – \\vec{p}_i}{\\Delta t} \\\\
\n &= \\frac{m\\vec{v}_f – m\\vec{v}_i}{\\Delta t}
\n\\end{align}
\n\\]<\/p>\n\n\n\n
\n\\begin{align*}
\n \\vec{F}_\\textit{net} &= \\frac{m\\left(\\vec{v}_f – \\vec{v}_i\\right)}{\\Delta t} \\\\
\n &= m\\frac{\\Delta\\vec{v}}{\\Delta t}
\n\\end{align*}
\n\\]<\/p>\n\n\n\n
\n\\vec{F}_\\textit{net} = m\\vec{a} \\tag{9.2}
\n\\]<\/p>\n\n\n\n
\n\\begin{align}
\n F_\\textit{net,x} &= \\frac{\\Delta p_x}{\\Delta t} & F_\\textit{net,y} &= \\frac{\\Delta p_y}{\\Delta t} \\tag{9.3} \\\\
\n F_\\textit{net,x} &= ma_x & F_\\textit{net,y} &= ma_y \\tag{9.4}
\n\\end{align}
\n\\]<\/p>\n\n\n\nExample 9.1<\/h4>\n\n\n\n
\n\\begin{align*}
\n F_\\textit{net,x} &= \\frac{\\Delta p_x}{\\Delta t} \\\\
\n &= m\\frac{\\Delta v}{\\Delta t} \\\\
\n &= m\\frac{v_f – v_i}{\\Delta t} \\\\
\n &= (1380\\ \\textrm{kg})\\left(\\frac{17 – 27\\ \\textrm{m\/s}}{8\\ \\textrm{s}}\\right) \\\\
\n &= -1725\\ \\textrm{N}
\n\\end{align*}
\n\\]<\/p>\n\n\n\nPractice 9.1<\/h4>\n\n\n\n\n