A

Step 1: Identify the forces acting on the cube.
When the copper cube slides down the inclined plane at a constant speed, the net force acting on it is zero. The forces involved are:
1. Gravitational force ($ F_g = mg $)
2. Normal force ($ F_N $)
3. Frictional force ($ F_f $)

Step 2: Calculate the gravitational force.
Given:
- Mass, $ m = 200 ext{ g} = 0.2 ext{ kg} $
- Gravitational acceleration, $ g = 9.8 ext{ m/s}^2 $

The gravitational force acting down the plane is:
\[ F_g = mg = 0.2 imes 9.8 = 1.96 ext{ N} \]

Step 3: Calculate the component of gravitational force along the incline.
The component of gravitational force along the incline:
\[ F_{g, ext{parallel}} = mg imes ext{sin}( heta) = 1.96 imes ext{sin}(37^ ext{o})\]
Using $ ext{sin}(37^ ext{o}) $ = 0.6:
\[ F_{g, ext{parallel}} = 1.96 imes 0.6 ext{ N} = 1.176 ext{ N} \]

Step 4: Calculate the frictional force.
Since the block moves at a constant speed, the frictional force ($ F_f $) must equal the component of gravitational force down the incline:
\[ F_f = F_{g, ext{parallel}} = 1.176 ext{ N} \]

Step 5: Calculate the work done by friction.
The work done by friction while sliding down 60 cm (or 0.6 m) is:
\[ W = F_f imes d = 1.176 ext{ N} imes 0.6 ext{ m} = 0.7056 ext{ J} \]

Step 6: Relate work done to thermal energy.
Assuming all the work done against friction turns into thermal energy, we can now find the increase in temperature of the copper block using its specific heat capacity.
\[ Q = m imes c imes riangle T \]
Where:
- $ Q $ = heat energy ($ 0.7056 ext{ J} $)
- $ c $ = specific heat capacity of copper = 420 J/kg-K
- $ riangle T $ = increase in temperature (K)

Rearranging to find $ riangle T $:
\[ riangle T = \frac{Q}{m imes c} = \frac{0.7056 ext{ J}}{0.2 ext{ kg} imes 420 ext{ J/kg-K}} \]
Substituting the values:
\[ riangle T = \frac{0.7056}{84} = 0.0084 ext{ K} \]

Final Step: Conclusion.
The increase in the temperature of the block as it slides down 60 cm is approximately 0.0084 K.