Certainly! Let's analyze the situation carefully to determine the relationship between the forces [tex]\( F_1 \)[/tex] and [tex]\( F_2 \)[/tex] in both scenarios.
### Given:
- [tex]\( F_1 \)[/tex]: Force applied downward at [tex]\( 25^\circ \)[/tex] below the horizontal.
- [tex]\( F_2 \)[/tex]: Force applied upward at [tex]\( 25^\circ \)[/tex] above the horizontal.
- The sled moves with the same acceleration in both cases, and friction is negligible.
### Analysis:
1. Horizontal Components of the Forces:
- When pushing with [tex]\( F_1 \)[/tex]:
[tex]\[
F_{1x} = F_1 \cdot \cos(25^\circ)
\][/tex]
- When pulling with [tex]\( F_2 \)[/tex]:
[tex]\[
F_{2x} = F_2 \cdot \cos(25^\circ)
\][/tex]
Since the sled moves with the same acceleration in both cases, the net horizontal force is the same:
[tex]\[
F_{1x} = F_{2x}
\][/tex]
Substituting the expressions for the horizontal components:
[tex]\[
F_1 \cdot \cos(25^\circ) = F_2 \cdot \cos(25^\circ)
\][/tex]
Given that [tex]\( \cos(25^\circ) \)[/tex] is a non-zero constant, we can divide both sides by [tex]\( \cos(25^\circ) \)[/tex]:
[tex]\[
F_1 = F_2
\][/tex]
So the correct statement is:
[tex]\[
\boxed{F_1 = F_2}
\][/tex]
### Explanation:
Despite the forces being applied in different directions relative to the horizontal (one downward and one upward), the horizontal component responsible for moving the sled forward remains the same in both cases due to identical acceleration and negligible friction. Therefore, the magnitudes of the forces must be equal.
