To determine the force \( F \) required to slide the stacked wood at constant speed across a horizontal surface, we can analyze the situation based on friction.
When the student first slides a single piece of wood with a force of 10 newtons at constant speed, it indicates that this applied force is equal to the frictional force resisting the motion. In this scenario, we assume that the coefficient of friction and the normal force are the only factors to consider.
When the piece of wood is cut in half and one piece is placed on top of the other, the total weight of the stacked wood increases.
Assuming \( W \) is the weight of the original piece of wood, the normal force acting on the stacked wood will be the combined weight of the two pieces. Therefore, the new normal force \( N \) becomes:
\[ N = 2W \]
The frictional force \( f \) can be calculated using the equation:
\[ f = \mu N \]
Where \( \mu \) is the coefficient of friction. Since the original piece of wood was moving at constant speed with a 10 Newton force, we can conclude that:
\[ 10 , \text{N} = \mu W \]
For the stacked wood, the frictional force becomes:
\[ f_{\text{stacked}} = \mu (2W) = 2 \mu W \]
Plugging in for \( f_{\text{stacked}} \):
\[ f_{\text{stacked}} = 2 \times 10 , \text{N} = 20 , \text{N} \]
Therefore, to slide the stacked wood at constant speed, the magnitude of the force \( F \) required is:
\[ \boxed{20 , \text{N}} \]
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