To find the normal force on the cart, we need to consider the forces acting on it. The cart has a weight due to gravity and is subject to the vertical component of the applied force.

1. Weight of the cart (W): $W = m \cdot g = 40 \, \text{kg} \cdot 9.81 \, \text{m/s}^2 = 392.4 \, \text{N}$

2. Vertical component of the applied force (F_v): The applied force is 200 N at an angle of 45° from the horizontal. $F_v = F \cdot \sin(\theta) = 200 \, \text{N} \cdot \sin(45^\circ) = 200 \, \text{N} \cdot \frac{\sqrt{2}}{2} \approx 141.42 \, \text{N}$

3. Normal force (N): The normal force acts upward, balancing the weight of the cart and the vertical component of the applied force. Thus, the normal force is given by: $N = W - F_v = 392.4 \, \text{N} - 141.42 \, \text{N} \approx 250.98 \, \text{N}$

The normal force on the cart is closest to 251 N.

Do you want details or have any questions?

Here are some related questions you might consider:

1. How does the angle of the applied force affect the normal force?
2. What would happen to the normal force if the angle increased to 60°?
3. How would friction change the normal force calculation?
4. What are the implications of a heavier cart on the normal force?
5. How do you calculate the net force acting on the cart horizontally?

Tip: Always resolve forces into their components when dealing with angled forces; it simplifies calculations significantly.