Question

a box is at rest on a table. the weight of the box is choose your answer... the magnitude of the tables upward force, and the weight vector and upward force vector are choose your answer... . (c) if the box were moving across the desk rightward a diagram would indicate (a) choose your answer... , (b) choose your answer... . choose you oose your answer... . choose your answer... greater than less than equal to

Explanation:

Step1: Analyze the forces on the box at rest

When the box is at rest on the table, it is in equilibrium (Newton's first law). The weight of the box (a downward force due to gravity) and the table's upward normal force are the two vertical forces acting on it. For an object in equilibrium (at rest or moving at constant velocity), the net force is zero. So, the magnitude of the weight (downward) and the upward normal force from the table must be equal in magnitude and opposite in direction.

Step2: Determine the relationship between weight and upward force

Since the box is at rest, the vertical forces (weight and upward force from the table) are balanced. By Newton's first law, the net force \( F_{net} = 0 \) in the vertical direction. Let \( W \) be the weight (downward) and \( N \) be the normal (upward) force. Then \( F_{net,y}=N - W=0\), so \( N = W \). So the magnitude of the weight of the box is equal to the magnitude of the table's upward force.

Step3: Analyze the forces when moving rightward at constant velocity (assuming constant velocity, as no acceleration is mentioned for the moving case)

If the box is moving rightward across the desk at a constant velocity, it is still in equilibrium (Newton's first law, since acceleration \( a = 0 \)). In the vertical direction, the weight and the upward normal force still balance (same as the rest case), so their magnitudes are equal. In the horizontal direction, if it's moving at constant velocity, the applied force (if any) and the frictional force balance, but the question here is about the vertical forces (weight and upward force) which still have equal magnitudes as in the rest case (because vertical motion isn't changing, so vertical net force is zero).

Answer:

For the first part (box at rest): The magnitude of the weight of the box is equal to the magnitude of the table's upward force.

For the second part (box moving rightward at constant velocity): The magnitude of the weight of the box is equal to the magnitude of the table's upward force.