Alright, let's carefully analyze each problem and provide detailed solutions.
### Problem 1:
Boy A and Boy B are pulling a heavy cabinet at the same time with 5 Newtons (N) of force each.
#### Solution for Problem 1:
- Since both boys are pulling in the same direction, the net force acting on the cabinet is the sum of their individual forces. Therefore:
[tex]\[
\text{Net force} = 5 \, \text{N} + 5 \, \text{N} = 10 \, \text{N}
\][/tex]
So, the net force acting on the cabinet is 10 N.
### Problem 2:
Boy A and Boy B pull the heavy cabinet at the same time in opposite directions. Boy A pulls with 10 N of force while Boy B pulls with 5 N of force.
#### Solution for Problem 2a:
- To find the net force, we take the difference between the forces since they are in opposite directions. Therefore:
[tex]\[
\text{Net force} = 10 \, \text{N} - 5 \, \text{N} = 5 \, \text{N}
\][/tex]
So, the net force on the cabinet is 5 N.
#### Solution for Problem 2b:
- Will the cabinet move? When there is a non-zero net force acting on an object, it will move in the direction of the net force. In this case, the net force is 5 N, which is not zero.
Therefore, the cabinet will move.
#### Solution for Problem 2c:
- Direction of movement: The direction of movement is determined by the direction of the net force. Since Boy A is pulling with a larger force (10 N) compared to Boy B (5 N), the net force direction will be toward Boy A’s side.
So, the cabinet will move in the direction of Boy A.
In summary:
1. The net force acting on the cabinet when both boys pull with 5 N each is 10 N.
2. When Boy A pulls with 10 N and Boy B pulls with 5 N in opposite directions:
- The net force on the cabinet is 5 N.
- The cabinet will move.
- The cabinet will move in the direction of Boy A.
