A rightward force is applied to a 6-kg object to move it across a rough surface at constant vel The object encounters 15 N of frictional force. Use the diagram to determine the gravitational force (weight), normal force, net force, and applied force. (Neglect air resistance.) ) FBD: b) Fweight: C)FN d) F net: e) Fa:

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Chapter1: Units, Trigonometry. And Vectors
Section: Chapter Questions
Problem 1CQ: Estimate the order of magnitude of the length, in meters, of each of the following; (a) a mouse, (b)...
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**Problem Statement:**

A rightward force is applied to a 6-kg object to move it across a rough surface at constant velocity. The object encounters 15 N of frictional force. Use the diagram to determine the gravitational force (weight), normal force, net force, and applied force. (Neglect air resistance.)

**Tasks:**
- a) Draw and label the Free Body Diagram (FBD) to illustrate the forces acting on the object.
- b) Determine the gravitational force (\(F_{\text{weight}}\)).
- c) Determine the normal force (\(F_{\text{N}}\)).
- d) Determine the net force (\(F_{\text{net}}\)).
- e) Determine the applied force (\(F_{\text{a}}\)).

**Note:** The question implies the necessity of analyzing different forces, such as gravitational force, normal force, and friction. Given that the object moves at a constant velocity, the net force should be zero, implying that the applied force should balance the frictional force.

**Diagram Explanation:**

While the image does not provide a diagram, the following components should be included in a properly drawn FBD for this scenario:

- **Gravitational Force (\(F_{\text{weight}}\))**: Acts downward, equal to the mass (\(m\)) times the acceleration due to gravity (\(g\)). For a 6-kg object, \(F_{\text{weight}} = 6 \times 9.8 = 58.8 \, \text{N}\).
  
- **Normal Force (\(F_{\text{N}}\))**: Acts perpendicular to the surface, equal and opposite to the gravitational force in magnitude since there is no vertical acceleration. Thus, \(F_{\text{N}} = 58.8 \, \text{N}\).
  
- **Frictional Force (\(F_{\text{friction}}\))**: Acts to the left, given in the problem as 15 N.
  
- **Applied Force (\(F_{\text{a}}\))**: Acts to the right, equal in magnitude to the frictional force for constant velocity, thus \(F_{\text{a}} = 15 \, \text{N}\).
  
- **Net Force (\(F_{\text{net}}\))**: Since the object moves at a constant velocity, \(F_{\text{net}} = 0 \
Transcribed Image Text:**Problem Statement:** A rightward force is applied to a 6-kg object to move it across a rough surface at constant velocity. The object encounters 15 N of frictional force. Use the diagram to determine the gravitational force (weight), normal force, net force, and applied force. (Neglect air resistance.) **Tasks:** - a) Draw and label the Free Body Diagram (FBD) to illustrate the forces acting on the object. - b) Determine the gravitational force (\(F_{\text{weight}}\)). - c) Determine the normal force (\(F_{\text{N}}\)). - d) Determine the net force (\(F_{\text{net}}\)). - e) Determine the applied force (\(F_{\text{a}}\)). **Note:** The question implies the necessity of analyzing different forces, such as gravitational force, normal force, and friction. Given that the object moves at a constant velocity, the net force should be zero, implying that the applied force should balance the frictional force. **Diagram Explanation:** While the image does not provide a diagram, the following components should be included in a properly drawn FBD for this scenario: - **Gravitational Force (\(F_{\text{weight}}\))**: Acts downward, equal to the mass (\(m\)) times the acceleration due to gravity (\(g\)). For a 6-kg object, \(F_{\text{weight}} = 6 \times 9.8 = 58.8 \, \text{N}\). - **Normal Force (\(F_{\text{N}}\))**: Acts perpendicular to the surface, equal and opposite to the gravitational force in magnitude since there is no vertical acceleration. Thus, \(F_{\text{N}} = 58.8 \, \text{N}\). - **Frictional Force (\(F_{\text{friction}}\))**: Acts to the left, given in the problem as 15 N. - **Applied Force (\(F_{\text{a}}\))**: Acts to the right, equal in magnitude to the frictional force for constant velocity, thus \(F_{\text{a}} = 15 \, \text{N}\). - **Net Force (\(F_{\text{net}}\))**: Since the object moves at a constant velocity, \(F_{\text{net}} = 0 \
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