Answer to Question #116030 in Classical Mechanics for kola Babs

Question #116030
an elastic material has a length of 36cm when a load of 40N is hung on it and a length of 45cm when a load of 60N is hung on it. what is the original length of the screen?
1
Expert's answer
2020-05-18T09:59:08-0400

For Hooke's law "F = Ke"


"\\frac Fe=K"

"\\frac{f_1}{e_1}=\\frac{f_2}{e_2}"

Let the original length=t0


therefore "e_1=(36\u2212t_0)" cm; "e_2=(46\u2212t_0)" cm

if "f_1=40N\\ \\& \\ f_2=60N"


"Then \\ \\frac{40}{\n\n36\u2212t_0}=\\frac{60}{(45\u2212t_0)}"

"40(45\u2212t_0)=60(36\u2212t_0)"

therefore "1800\u221240t_0 = 2160\u221260t_0"

"60t_0\u221240t_0=2160\u22121800"

"20t_0 = 360"

"t_0 = 18cm"



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