Answer:
a) k = 3.6 N/m ; b) 9.43 * 10¹²; c) 1.18 * 10¹¹; d) 75.08 N/m
Step-by-step explanation:
Length of iron bar = L = 2.7m;
side length of cross section = a = 0.07cm = 0.0007m;
x = 2.7 cm = 0.027m;
m = 100kg;
ρ = 7.87gm/cm³;
da = 2.28 * 10⁻¹⁰m;
a)
Fnet = F - mg
where Fnet = 0
So,
F = mg where F=kx
k = mg/x = 100*9.8/0.027
k = 3.6 N/m
b)
Nchain = Aw/Aa =a²/(da)²
= (o.ooo7)²/(2.28 * 10⁻¹⁰)²
= 9.43 * 10¹²
c)
Nbond = L/da = 2.7/2.28 * 10⁻¹⁰
= 1.18 * 10¹¹
d)
Spring stiffness of wire = ksi = (Nbondk)/Nchain
= [(1.18* 10¹¹)(6*10⁴)]/(9.43 * 10¹²)
= 75.08 N/m
