Question

the free body diagram shows a box being pulled to the left up a 25 - degree incline. the magnitude of the normal force is n. 20 45 50 75

Explanation:

Step1: Analyze vertical - force equilibrium

In the vertical direction, assume the normal force is $N$. The vertical - component of the $45N$ force is $45\sin25^{\circ}$, and the vertical forces acting on the box are the downward $50N$ force and the upward vertical - component of the $45N$ force and the normal force $N$. In equilibrium, the sum of vertical forces is zero.

Step2: Set up the force - equilibrium equation

$\sum F_y = 0$. So, $N+45\sin25^{\circ}-50 = 0$. First, calculate $45\sin25^{\circ}\approx45\times0.4226 = 19.017\approx20N$. Then, $N=50 - 45\sin25^{\circ}\approx50 - 20=30N$. But if we assume there is no acceleration in the vertical direction and we consider the forces in a more general way, and if we assume the forces are already balanced in the context of the problem without considering the incline's effect on the normal force in a more complex way (since the problem seems to lack some information about the mass and gravity, and if we just consider the given forces in a simple force - balance in the vertical direction), and we note that the vertical forces acting on the point are a downward $50N$ and an upward component from the $45N$ force. If we assume the normal force balances the net vertical force, and considering the closest value from the given options, we assume the normal force is calculated based on the vertical - force balance of the given forces. The upward component of the $45N$ force is approximately $20N$. So the normal force $N = 50 - 20=30N$ (closest to the options analysis). But if we assume the problem is asking for a simple balance of the vertical forces among the given ones, and we take into account the options, we consider the vertical forces: a downward force of $50N$ and an upward component of the $45N$ force. The normal force $N$ that balances the vertical forces is such that if we assume the upward component of the $45N$ force is about $20N$ (from $45\sin25^{\circ}\approx20N$), then $N = 50 - 20 = 30N$. Since the closest option to $30N$ among the given ones is $20N$ (assuming some approximation or simplification in the problem setup).

Answer:

20