Answer:
The mass is [tex]m = 3.45408 kg[/tex]
Explanation:
From the question we are told that
The extension of the rod is [tex]from , \ x_1 = 0 \to x_2 = 13.0[/tex]
The area is [tex]A = 8.0 cm^2[/tex]
The density increase as follows [tex]from \ \rho_1 =2.5 g/cm^2 \to \rho_2= 19.0 g/cm^3[/tex]
The equation [tex]\rho = B + Cx[/tex]
at [tex]x_1= 0[/tex] [tex]\rho_1 =2.5 g/cm^2[/tex]
So
[tex]2.5 = B + 0[/tex]
=> [tex]B =2.5[/tex]
So at [tex]x_2 = 13.0[/tex] , [tex]\rho_2= 19.0 g/cm^3[/tex]
So
[tex]19.0 = 2.5 + C(13)[/tex]
=> [tex]C = 1.27[/tex]
Now
[tex]m = 8 \int\limits^{13}_{0} {2.5 + 1.27x} \, dx[/tex]
[tex]m = 8 [{2.5 +\frac{ 1.27x^2}{2} } ]\left | 13} \atop {0}} \right.[/tex]
[tex]m = 8 [{2.5 +\frac{ 1.27(13)^2}{2} } ][/tex]
[tex]m = 3454.08 g[/tex]
[tex]m = 3.45408 kg[/tex]
