Answer
The other end of the system of rods moves $2.205~mm$
Work Step by Step
We can find the increase in length of the copper rod when it is heated:
$\Delta L = \alpha~\Delta T~L$
$\Delta L = (17\times 10^{-6}~K^{-1})(150~K)(0.35~m)$
$\Delta L = 892.5\times 10^{-6}~m$
We can find the increase in length of the aluminum rod when it is heated:
$\Delta L = \alpha~\Delta T~L$
$\Delta L = (25\times 10^{-6}~K^{-1})(150~K)(0.35~m)$
$\Delta L = 1312.5\times 10^{-6}~m$
We can find the total increase in length of the system:
$\Delta L_{total} = 892.5\times 10^{-6}~m+1312.5\times 10^{-6}~m$
$\Delta L_{total} = 2205\times 10^{-6}~m$
$\Delta L_{total} = 2.205\times 10^{-3}~m$
$\Delta L_{total} = 2.205~mm$
The other end of the system of rods moves $2.205~mm$
