Answer
We can rank the situations according to the increase in thermal energy due to the sliding:
$(2) \gt (1) \gt (3)$
Work Step by Step
We can write an expression for the thermal energy:
$E_{th}+\Delta K + \Delta U_g = 0$
$E_{th} = -\Delta K - \Delta U_g$
Let the initial kinetic energy be $K$.
We can find an expression for the thermal energy in each situation:
Situation (1)
$E_{th} = -\Delta K - \Delta U_g = -(-K)-0 = K$
Situation (2)
$E_{th} = -\Delta K - \Delta U_g = -(-K)-\Delta U_g = K-\Delta U_g$
Since $\Delta U_g \lt 0,$ then $~~(K-\Delta U_g) \gt K$
Then $~~E_{th} \gt K$
Situation (3)
$E_{th} = -\Delta K - \Delta U_g = -(-K)-\Delta U_g = K-\Delta U_g$
Since $\Delta U_g \gt 0,$ then $~~(K-\Delta U_g) \lt K$
Then $~~E_{th} \lt K$
We can rank the situations according to the increase in thermal energy due to the sliding:
$(2) \gt (1) \gt (3)$
