A block is displaced vertically by Ar = 2.5 m under the influence of three forces, F = 4.0 N, FB = 6.0N, and Fc = 8.0N as shown in the diagram. The angles 0c = 15°. There are no other forces acting on the block. What is the net work done by these three forces on the block? FB %3D %3D %3D FA Oc W = FArcos(0) %3D

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**Understanding Work Done by Forces on a Block**

A block is displaced vertically by \(\Delta \vec{r} = 2.5 \, \text{m}\) under the influence of three forces. The forces are:

- \(F_A = 4.0 \, \text{N}\)
- \(F_B = 6.0 \, \text{N}\)
- \(F_C = 8.0 \, \text{N}\)

These are shown in the accompanying diagram. The angle \(\theta_C\) is \(15^\circ\). There are no other forces acting on the block. The task is to find the net work done by these forces on the block.

**Equation for Work**

The work done by a force \(F\) in a given displacement \(\Delta \vec{r}\) at an angle \(\theta\) is given by the formula:  

\[ W = F \Delta r \cos(\theta) \]

**Diagram Explanation**

- **Forces:**
  - \(F_A\) is directed horizontally to the left.
  - \(F_B\) is directed vertically upward.
  - \(F_C\) is directed diagonally downward to the right at an angle \(\theta_C\) with the vertical displacement.

- **Displacement:** 
  - Represented by a vertical arrow pointing upward, labeled \(\Delta \vec{r}\).

- **Angle \(\theta_C\):**
  - The angle between force \(F_C\) and the vertical displacement \(\Delta \vec{r}\) is \(15^\circ\).

The question asks for the calculation of the net work done by these forces based on the given conditions and the provided work formula.
Transcribed Image Text:**Understanding Work Done by Forces on a Block** A block is displaced vertically by \(\Delta \vec{r} = 2.5 \, \text{m}\) under the influence of three forces. The forces are: - \(F_A = 4.0 \, \text{N}\) - \(F_B = 6.0 \, \text{N}\) - \(F_C = 8.0 \, \text{N}\) These are shown in the accompanying diagram. The angle \(\theta_C\) is \(15^\circ\). There are no other forces acting on the block. The task is to find the net work done by these forces on the block. **Equation for Work** The work done by a force \(F\) in a given displacement \(\Delta \vec{r}\) at an angle \(\theta\) is given by the formula: \[ W = F \Delta r \cos(\theta) \] **Diagram Explanation** - **Forces:** - \(F_A\) is directed horizontally to the left. - \(F_B\) is directed vertically upward. - \(F_C\) is directed diagonally downward to the right at an angle \(\theta_C\) with the vertical displacement. - **Displacement:** - Represented by a vertical arrow pointing upward, labeled \(\Delta \vec{r}\). - **Angle \(\theta_C\):** - The angle between force \(F_C\) and the vertical displacement \(\Delta \vec{r}\) is \(15^\circ\). The question asks for the calculation of the net work done by these forces based on the given conditions and the provided work formula.
Expert Solution
Step 1

The force of the block is,

FB=FBj^=6.0 Nj^FA=-FAi^=-4.0 Ni^

The force acting along the point C is,

FC=FCcosθCi^-FCsinθCj^=8.0 Ncos15°i^-8.0 Nsin15°j^=7.7 Ni^-2.1 Nj^

The net force is,

F=FA+FB+FC=-4.0i^+6.0j^+7.7i^-2.1j^=3.7i^+3.9j^

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