During a move, you and two friends decide to move a heavy dresser. You pull the dresser to the right with force F,= 4.93 N. One useless friend leans on the dresser with a downward force of F,= 3.27 N. The other friend pulls with a force of F,= 8.58 N at an angle of 60 degrees as indicated in the image below. If the three of you manage to slide the dresser 3.24 m towards the right, what is the net work done on the dresser by the three forces? y F2

Physics for Scientists and Engineers: Foundations and Connections
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Author:Katz, Debora M.
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Chapter5: Newton's Laws Of Motion
Section: Chapter Questions
Problem 72PQ: A block of ice (m = 15.0 kg) with an attached rope is at rest on a frictionless surface. You pull...
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**Problem Description:**

During a move, you and two friends decide to move a heavy dresser. You pull the dresser to the right with a force \( F_1 = 4.93 \, \text{N} \). One friend applies a downward force of \( F_2 = 3.27 \, \text{N} \) on the dresser. The other friend pulls with a force of \( F_3 = 8.58 \, \text{N} \) at an angle of 60 degrees, as depicted in the diagram. If the three of you manage to slide the dresser 3.24 m towards the right, what is the net work done on the dresser by the three forces?

**Diagram Explanation:**

- The diagram shows a coordinate system with the \( x \)-axis pointing to the right and the \( y \)-axis pointing upward.
- A brown rectangular block represents the dresser.
- Force \( F_1 \) is shown as an arrow pointing to the right along the \( x \)-axis.
- Force \( F_2 \) is a downward arrow aligned with the \( -y \)-axis.
- Force \( F_3 \) is an arrow positioned at a 60-degree angle from the horizontal axis, pointing upwards and to the left.
- The displacement \( \Delta x \) is indicated to the right along the \( x \)-axis.

**Calculation of Net Work Done:**

To calculate the net work done by these forces, we need to consider only the components of the forces that are parallel to the direction of displacement (\( \Delta x \)).

1. **Work done by \( F_1 \):**
   \[
   W_1 = F_1 \cdot \Delta x = 4.93 \, \text{N} \times 3.24 \, \text{m} = 15.98 \, \text{J}
   \]

2. **Work done by \( F_2 \):**
   - Since \( F_2 \) is perpendicular to the direction of motion, it does no work.
   \[
   W_2 = 0
   \]

3. **Work done by \( F_3 \):**
   - The horizontal component of \( F_3 \) is \( F_{3x} = F_3 \cos(60^\circ) \).
Transcribed Image Text:**Problem Description:** During a move, you and two friends decide to move a heavy dresser. You pull the dresser to the right with a force \( F_1 = 4.93 \, \text{N} \). One friend applies a downward force of \( F_2 = 3.27 \, \text{N} \) on the dresser. The other friend pulls with a force of \( F_3 = 8.58 \, \text{N} \) at an angle of 60 degrees, as depicted in the diagram. If the three of you manage to slide the dresser 3.24 m towards the right, what is the net work done on the dresser by the three forces? **Diagram Explanation:** - The diagram shows a coordinate system with the \( x \)-axis pointing to the right and the \( y \)-axis pointing upward. - A brown rectangular block represents the dresser. - Force \( F_1 \) is shown as an arrow pointing to the right along the \( x \)-axis. - Force \( F_2 \) is a downward arrow aligned with the \( -y \)-axis. - Force \( F_3 \) is an arrow positioned at a 60-degree angle from the horizontal axis, pointing upwards and to the left. - The displacement \( \Delta x \) is indicated to the right along the \( x \)-axis. **Calculation of Net Work Done:** To calculate the net work done by these forces, we need to consider only the components of the forces that are parallel to the direction of displacement (\( \Delta x \)). 1. **Work done by \( F_1 \):** \[ W_1 = F_1 \cdot \Delta x = 4.93 \, \text{N} \times 3.24 \, \text{m} = 15.98 \, \text{J} \] 2. **Work done by \( F_2 \):** - Since \( F_2 \) is perpendicular to the direction of motion, it does no work. \[ W_2 = 0 \] 3. **Work done by \( F_3 \):** - The horizontal component of \( F_3 \) is \( F_{3x} = F_3 \cos(60^\circ) \).
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