Answer
(a) $0.04$
(b) $0.04$
(c) $0.046$
Work Step by Step
(a) We can find the fractional increase in length as follows:
$F=mg$
$F=(4.8Kg)(9.8m/s^2)=47.04N$
and $A=\frac{\pi}{4}D^2$
$A=\frac{\pi}{4}(0.54\times 10^{-3}m)^2=0.23\times 10^{-6}m^2$
Now $\frac{\Delta L_{\circ}}{L_{\circ}}=\frac{F}{YA}$
We plug in the known values to obtain:
$\frac{\Delta L_{\circ}}{L_{\circ}}=\frac{47.04Kg.m/s^2}{5.1\times 10^9N/m^2(0.23\times 10^{-6}m^2)}$
$\frac{\Delta L_{\circ}}{L_{\circ}}=0.04$
(b) We know that when the fishing line is pulled at constant speed then the acceleration and force is zero. As a result, the fractional increase in length will remain the same (that is, $0.04$).
(c) We know that
$\frac{\Delta L_{\circ}}{L_{\circ}}=\frac{F}{YA}$
We plug in the known values to obtain:
$\frac{\Delta L_{\circ}}{L_{\circ}}=\frac{48Kg(1.5m/s^2+9.81m/s^2)}{5.1\times 10^9N/m^2(0.23\times 10^{-6}m^2)}$
$\frac{\Delta L_{\circ}}{L_{\circ}}=0.046$
