College Physics
11th Edition
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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![**Problem Statement:**
A box is subjected to multiple forces. The forces involved are as follows:
- \( F_N = 100 \, \text{N} \) (normal force, pointing upwards)
- \( F_g = 100 \, \text{N} \) (gravitational force, pointing downwards)
- \( F_f = 90 \, \text{N} \) (force of friction, pointing left)
- \( F_a = 120 \, \text{N} \) (applied force, pointing right)
**Question:**
If the mass of the box is 15 kg, what is its acceleration?
**Answer Options:**
A) \( 2 \, \text{m/s}^2 \)
B) \( 450 \, \text{m/s}^2 \)
C) \( 0.5 \, \text{m/s}^2 \)
D) \( 90 \, \text{m/s}^2 \)
**Diagram Explanation:**
- The diagram shows a square representing the box.
- Arrows indicate the direction and magnitude of each force.
- \( F_N \) (100 N) is upwards, opposing gravitational force \( F_g \) (100 N), indicating vertical forces are balanced.
- \( F_f \) (90 N) is to the left, representing friction.
- \( F_a \) (120 N) is to the right, representing the applied force.
**Analysis:**
Horizontal Force Calculation:
- Net force (\( F_{\text{net}} \)) in the horizontal direction is calculated by subtracting the frictional force from the applied force:
\[
F_{\text{net}} = F_a - F_f = 120 \, \text{N} - 90 \, \text{N} = 30 \, \text{N}
\]
Acceleration Calculation:
- Using Newton’s second law: \( F = ma \)
\[
a = \frac{F_{\text{net}}}{m} = \frac{30 \, \text{N}}{15 \, \text{kg}} = 2 \, \text{m/s}^2
\]
**Correct Answer:**
A) \( 2 \, \text{m/s}^2 \)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fa36fb3a7-f72c-48bd-9564-279377522d6c%2F15d9948a-e459-4afd-9187-844f8f4792bd%2F943maa9_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A box is subjected to multiple forces. The forces involved are as follows:
- \( F_N = 100 \, \text{N} \) (normal force, pointing upwards)
- \( F_g = 100 \, \text{N} \) (gravitational force, pointing downwards)
- \( F_f = 90 \, \text{N} \) (force of friction, pointing left)
- \( F_a = 120 \, \text{N} \) (applied force, pointing right)
**Question:**
If the mass of the box is 15 kg, what is its acceleration?
**Answer Options:**
A) \( 2 \, \text{m/s}^2 \)
B) \( 450 \, \text{m/s}^2 \)
C) \( 0.5 \, \text{m/s}^2 \)
D) \( 90 \, \text{m/s}^2 \)
**Diagram Explanation:**
- The diagram shows a square representing the box.
- Arrows indicate the direction and magnitude of each force.
- \( F_N \) (100 N) is upwards, opposing gravitational force \( F_g \) (100 N), indicating vertical forces are balanced.
- \( F_f \) (90 N) is to the left, representing friction.
- \( F_a \) (120 N) is to the right, representing the applied force.
**Analysis:**
Horizontal Force Calculation:
- Net force (\( F_{\text{net}} \)) in the horizontal direction is calculated by subtracting the frictional force from the applied force:
\[
F_{\text{net}} = F_a - F_f = 120 \, \text{N} - 90 \, \text{N} = 30 \, \text{N}
\]
Acceleration Calculation:
- Using Newton’s second law: \( F = ma \)
\[
a = \frac{F_{\text{net}}}{m} = \frac{30 \, \text{N}}{15 \, \text{kg}} = 2 \, \text{m/s}^2
\]
**Correct Answer:**
A) \( 2 \, \text{m/s}^2 \)
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