e Fat an angle to the Ar cos 8. positive, so the sign of Wis force and the displacement. 1 of 1 What is the work W., done on the box by the normal force? Express your answers in joules to two significant figures. ▸ View Available Hint(s) W₁ = 0 J Submit Correct Because there is no component of the normal force in the direction of motion, the normal ford Part C Previous Answers What is the work WT done by the tension force? Express your answers in joules to two significant figures. ▸ View Available Hint(s) WT= Submit 195| ΑΣΦΑ Previous Answers J * Incorrect; Try Again; 5 attempts remaining Since the force of friction acts to oppose the motion of the box, does it add energy to the box force does negative work on the system.

College Physics
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ISBN:9781305952300
Author:Raymond A. Serway, Chris Vuille
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Chapter1: Units, Trigonometry. And Vectors
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Part C

### Physics Problem Involving an Inclined Plane

**Problem Statement:**

A box with a weight of magnitude \( F_G = 2.00 \, \text{N} \) is lowered by a rope down a smooth plane inclined at an angle \( \phi = 30.0^\circ \) above the horizontal. This setup is depicted in Figure 1. The normal force acting on the box has a magnitude \( n = 1.73 \, \text{N} \). The tension force in the rope is \( 1.00 \, \text{N} \), and the displacement \( \Delta \vec{r} \) of the box is \( 1.80 \, \text{m} \) down the inclined plane.

**Diagram Explanation:**

Figure 1 (not shown here) likely illustrates the inclined plane with the box positioned on it. The plane is inclined at 30 degrees to the horizontal, demonstrating the direction of forces acting on the box:

- **Gravitational force (\( F_G \))** is directed vertically downwards.
- **Normal force (\( n \))** is perpendicular to the surface of the inclined plane.
- **Tension force** is directed along the plane, in the direction of the rope.
- **Displacement (\( \Delta \vec{r} \))** is down the incline, showing the path of the box.

These forces and vectors may be represented with arrows to indicate their directions and relative magnitudes.
Transcribed Image Text:### Physics Problem Involving an Inclined Plane **Problem Statement:** A box with a weight of magnitude \( F_G = 2.00 \, \text{N} \) is lowered by a rope down a smooth plane inclined at an angle \( \phi = 30.0^\circ \) above the horizontal. This setup is depicted in Figure 1. The normal force acting on the box has a magnitude \( n = 1.73 \, \text{N} \). The tension force in the rope is \( 1.00 \, \text{N} \), and the displacement \( \Delta \vec{r} \) of the box is \( 1.80 \, \text{m} \) down the inclined plane. **Diagram Explanation:** Figure 1 (not shown here) likely illustrates the inclined plane with the box positioned on it. The plane is inclined at 30 degrees to the horizontal, demonstrating the direction of forces acting on the box: - **Gravitational force (\( F_G \))** is directed vertically downwards. - **Normal force (\( n \))** is perpendicular to the surface of the inclined plane. - **Tension force** is directed along the plane, in the direction of the rope. - **Displacement (\( \Delta \vec{r} \))** is down the incline, showing the path of the box. These forces and vectors may be represented with arrows to indicate their directions and relative magnitudes.
**Learning Goal:**
Recall that the work \( W \) done by a constant force \( \vec{F} \) at an angle \( \theta \) to the displacement \( \vec{\Delta r} \) is 

\[ W = \vec{F} \cdot \vec{\Delta r} = F \Delta r \cos \theta. \]

The vector magnitudes \( F \) and \( \Delta r \) are always positive, so the sign of \( W \) is determined entirely by the angle \( \theta \) between the force and the displacement.

**Figure:**
- The diagram illustrates a box being pulled up along an inclined plane.
- The box is subjected to two forces: a tension force \( \vec{T} \) along the incline and friction \( \vec{f} \) opposing the motion.
- The displacement \( \vec{\Delta r} \) is shown parallel to the incline.

---

**Part B**

What is the work \( W_n \) done on the box by the normal force?

- **Express your answer in joules to two significant figures.**

\[ W_n = 0 \, J \]

- **Correct**
  - Because there is no component of the normal force in the direction of motion, the normal force does no work on the box.

---

**Part C**

What is the work \( W_T \) done by the tension force?

- **Express your answer in joules to two significant figures.**

\[ W_T = \]

- **Incorrect; Try Again; 5 attempts remaining**
  - Since the force of friction acts to oppose the motion of the box, does it add energy to the box or take energy away from the box? When a force acts to remove energy from a system, that force does negative work on the system.
Transcribed Image Text:**Learning Goal:** Recall that the work \( W \) done by a constant force \( \vec{F} \) at an angle \( \theta \) to the displacement \( \vec{\Delta r} \) is \[ W = \vec{F} \cdot \vec{\Delta r} = F \Delta r \cos \theta. \] The vector magnitudes \( F \) and \( \Delta r \) are always positive, so the sign of \( W \) is determined entirely by the angle \( \theta \) between the force and the displacement. **Figure:** - The diagram illustrates a box being pulled up along an inclined plane. - The box is subjected to two forces: a tension force \( \vec{T} \) along the incline and friction \( \vec{f} \) opposing the motion. - The displacement \( \vec{\Delta r} \) is shown parallel to the incline. --- **Part B** What is the work \( W_n \) done on the box by the normal force? - **Express your answer in joules to two significant figures.** \[ W_n = 0 \, J \] - **Correct** - Because there is no component of the normal force in the direction of motion, the normal force does no work on the box. --- **Part C** What is the work \( W_T \) done by the tension force? - **Express your answer in joules to two significant figures.** \[ W_T = \] - **Incorrect; Try Again; 5 attempts remaining** - Since the force of friction acts to oppose the motion of the box, does it add energy to the box or take energy away from the box? When a force acts to remove energy from a system, that force does negative work on the system.
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