\nThe following is by my understanding.<\/p>\n
\nHooke’s Law is often visited in Physics 11, and applies to springs. It can be used to find a spring constant, the force involved to compress or stretch a spring to a given new length, or how much a spring will change length when subject to a force.<\/p>\n
\nThe main formula connected with Hooke’s Law:<\/p>\n
\nF=-kx<\/p>\n
\nThe negative sign means the spring’s reaction force is against the force to change its length. The k symbol stands for the spring constant, given in force\/length, N\/m in base units. The x symbol stands for the change in length relative to rest length. F means the force in play.<\/p>\n
\nExample: imagine a spring has constant 55N\/m and a rest length of 40.cm. A 500.g mass is hung from it. Afterwards, what is its new length?<\/p>\n
\nSolution: 500.g=0.500kg, so gravitational force on it will be mg=0.500(-9.8)=-4.9N (the negative means downward). F=-kx\u21d2-F\/k=x\u21d2-(-4.9)\/55=0.089m or 8.9cm.<\/p>\n
\nTherefore, the spring will be stretched 8.9cm by the mass, resulting in its new length being 48.9cm after the mass is hung from it.<\/p>\n
\nThere is more to say about Hooke’s Law; I hope to follow up about it in coming posts:)<\/p>\n
\nSource:<\/p>\n
Heath,R.W., Macnaughton,R.R., Martindale,D.G. (1979). Fundamentals of Physics.<\/em> D.C. Heath Canada Ltd.<\/p>\nJack of Oracle Tutoring by Jack and Diane,<\/a> Campbell River, BC.\n","protected":false},"excerpt":{"rendered":" Tutoring physics, Hooke’s Law is perennial. The tutor begins about it. The following is by my …<\/p>\n