Question

13 the diagram below represents a block sliding across a table at a constant speed. all forces are shown. what is the magnitude of the net force on the block? a 2.5 n b 6.7 n c 1.7 n d 0.8 n

Explanation:

Step1: Analyze horizontal forces

Horizontal forces: \( F_{right} = 2.5\,\text{N} \), \( F_{left} = 2.5\,\text{N} \). Net horizontal force \( F_{net,x} = 2.5 - 2.5 = 0\,\text{N} \).

Step2: Analyze vertical force

Vertical force: Only \( F_{down} = 1.7\,\text{N} \) (no upward vertical force shown, but wait—wait, actually, the block is moving at constant speed, so net force should be zero? Wait, no, the diagram: horizontal forces cancel. Wait, maybe I misread. Wait, the vertical force: is there a normal force? Wait, no, the diagram shows horizontal forces (2.5N right, 2.5N left) and a downward 1.7N. Wait, but for constant speed, net force is zero. Wait, maybe the vertical force is friction? No, wait, the problem: the block is sliding at constant speed, so net force is zero? But the options include 1.7N? Wait, no, maybe I messed up. Wait, no—wait, the horizontal forces are equal (2.5N right and 2.5N left), so they cancel. The vertical force: is there a force? Wait, the diagram: the downward force is 1.7N, but is there an upward force? Wait, maybe the diagram is showing horizontal (friction and applied) and vertical (weight and normal)? No, the diagram as given: horizontal arrows (2.5N right, 2.5N left) and a downward 1.7N. Wait, maybe the 1.7N is the net force? Wait, no—wait, horizontal forces cancel (2.5 - 2.5 = 0), so net force is equal to the vertical force? Wait, no, that doesn't make sense. Wait, maybe the diagram is misinterpreted. Wait, the problem says "all forces are shown". So horizontal: 2.5N right (applied), 2.5N left (friction). Vertical: 1.7N down (maybe weight, but no normal? No, that can't be. Wait, no—wait, the block is on a table, so normal force should balance weight, but in the diagram, only 1.7N down. Wait, maybe the 1.7N is the net force? Wait, no, horizontal forces cancel, so net force is 1.7N? But the options have C as 1.7N. Wait, let's recalculate:

Net force is the vector sum. Horizontal components: \( F_x = 2.5 - 2.5 = 0 \). Vertical component: \( F_y = 1.7\,\text{N} \) (since no upward force shown? Wait, that can't be. Wait, maybe the diagram is showing friction (2.5N left), applied (2.5N right), and a force of 1.7N downward (maybe air resistance? No, sliding on table. Wait, maybe the problem has a typo, but according to the diagram, horizontal forces cancel, so net force is equal to the vertical force, which is 1.7N. So the net force is 1.7N.

Answer:

C. 1.7 N