**Problem Statement:**A rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30° with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N. **Questions:**1. How much work does each of the following forces do on the box? - (a) Gravity - (b) The tension in the rope - (c) Friction - (d) The normal force2. What is the total work done on the box?**Details for Explanation:**To solve for the work done by each force:- Calculate the components of each force along the direction of the motion (horizontal direction).- Use the work formula: \( W = F \cdot d \cdot \cos(\theta) \), where \( W \) is work, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and the direction of motion.**(a) Work done by gravity:**- Gravity acts vertically downward with force \( F = mg \).- Since the motion is horizontal, the angle \( \theta \) between gravity and the direction of motion is \( 90^\circ \).- Hence, work done by gravity \( W_g = F \cdot d \cdot \cos(90^\circ) = 0 \).**(b) Work done by the tension in the rope:**- The tension force is \( 5 \) N and makes an angle of \( 30^\circ \) with the horizontal.- The horizontal component of the tension: \( 5 \cdot \cos(30^\circ) = 5 \cdot (\sqrt{3}/2) = 4.33 \) N.- Work done by tension: \( W_t = 4.33 \cdot 1.5 = 6.495 \) J.**(c) Work done by friction:**- Friction force is \( 1 \) N opposing the motion.- Hence, the angle \( \theta \) between friction and the direction of motion is \( 180^\circ \).- Work done by friction: \( W_f = 1 \cdot 1.5 \cdot

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A rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30° with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N. How much work does each of the following forces do on the box: (a) gravity, (b) the tension in the rope, (c) friction, and (d) the normal force? What is the total work done on the box?

A rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30° with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N. How much work does each of the following forces do on the box: (a) gravity, (b) the tension in the rope, (c) friction, and (d) the normal force? What is the total work done on the box?

College Physics

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ISBN:9781305952300

Author:Raymond A. Serway, Chris Vuille

Publisher:Raymond A. Serway, Chris Vuille

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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Question

**Problem Statement:**

A rope is tied to a box and used to pull the box 1.5 m along a horizontal floor. The rope makes an angle of 30° with the horizontal and has a tension of 5 N. The opposing friction force between the box and the floor is 1 N.

**Questions:**

1. How much work does each of the following forces do on the box?
- (a) Gravity
- (b) The tension in the rope
- (c) Friction
- (d) The normal force

2. What is the total work done on the box?

**Details for Explanation:**

To solve for the work done by each force:
- Calculate the components of each force along the direction of the motion (horizontal direction).
- Use the work formula: \( W = F \cdot d \cdot \cos(\theta) \),
where \( W \) is work, \( F \) is force, \( d \) is displacement, and \( \theta \) is the angle between the force and the direction of motion.

**(a) Work done by gravity:**
- Gravity acts vertically downward with force \( F = mg \).
- Since the motion is horizontal, the angle \( \theta \) between gravity and the direction of motion is \( 90^\circ \).
- Hence, work done by gravity \( W_g = F \cdot d \cdot \cos(90^\circ) = 0 \).

**(b) Work done by the tension in the rope:**
- The tension force is \( 5 \) N and makes an angle of \( 30^\circ \) with the horizontal.
- The horizontal component of the tension: \( 5 \cdot \cos(30^\circ) = 5 \cdot (\sqrt{3}/2) = 4.33 \) N.
- Work done by tension: \( W_t = 4.33 \cdot 1.5 = 6.495 \) J.

**(c) Work done by friction:**
- Friction force is \( 1 \) N opposing the motion.
- Hence, the angle \( \theta \) between friction and the direction of motion is \( 180^\circ \).
- Work done by friction: \( W_f = 1 \cdot 1.5 \cdot

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