A 1.20 kg object is subjected to three forces that give it an acceleration à = (4.80 m/s² )î + (5.00 m/s² )}. If two of the three : F₁ = (28.0 N )î + (11.0 N )Ĵ and F₂ = (15.0 N )î + (8.00 N )ĵ, find (a) the x component and (b) the y component of the the third force.
A 1.20 kg object is subjected to three forces that give it an acceleration à = (4.80 m/s² )î + (5.00 m/s² )}. If two of the three : F₁ = (28.0 N )î + (11.0 N )Ĵ and F₂ = (15.0 N )î + (8.00 N )ĵ, find (a) the x component and (b) the y component of the the third force.
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Chapter1: Units, Trigonometry. And Vectors
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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 Description:**
A 1.20 kg object is subjected to three forces that give it an acceleration \(\vec{a} = (4.80 \, \text{m/s}^2) \hat{\imath} + (5.00 \, \text{m/s}^2) \hat{\jmath}\). If two of the three forces are \(\vec{F}_1 = (28.0 \, \text{N}) \hat{\imath} + (11.0 \, \text{N}) \hat{\jmath}\) and \(\vec{F}_2 = (15.0 \, \text{N}) \hat{\imath} + (8.00 \, \text{N}) \hat{\jmath}\), find (a) the x component and (b) the y component of the third force.
**Explanation of Formulae and Notation:**
- **\(\vec{a}\)**: Represents the acceleration vector of the object, given in unit vector notation.
- **\(\vec{F}_1\) and \(\vec{F}_2\)**: Represent two out of the three forces acting on the object, given in unit vector notation (\(\hat{\imath}\) and \(\hat{\jmath}\) denote the x and y components, respectively).
- **Mass of the object**: 1.20 kg.
**Steps to Solve:**
1. **Apply Newton's Second Law**:
\[
\sum \vec{F} = m \vec{a}
\]
where \(m\) is the mass of the object, and \(\sum \vec{F}\) is the vector sum of all forces acting on the object.
2. **Calculate \(\sum \vec{F}\)**:
- Using the formula, \(m = 1.20 \, \text{kg}\) and \(\vec{a} = (4.80 \, \text{m/s}^2) \hat{\imath} + (5.00 \, \text{m/s}^2) \hat{\jmath}\).
\[
\sum \vec{F} = 1.20 \, \text{kg} \times \left[(4.80 \, \text{m/s](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F084886ad-eaf9-4e39-9ea7-4eb8cf5e9648%2Ff4880e09-3f68-4adf-bcde-a70afa773aad%2Fwyudly9_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Description:**
A 1.20 kg object is subjected to three forces that give it an acceleration \(\vec{a} = (4.80 \, \text{m/s}^2) \hat{\imath} + (5.00 \, \text{m/s}^2) \hat{\jmath}\). If two of the three forces are \(\vec{F}_1 = (28.0 \, \text{N}) \hat{\imath} + (11.0 \, \text{N}) \hat{\jmath}\) and \(\vec{F}_2 = (15.0 \, \text{N}) \hat{\imath} + (8.00 \, \text{N}) \hat{\jmath}\), find (a) the x component and (b) the y component of the third force.
**Explanation of Formulae and Notation:**
- **\(\vec{a}\)**: Represents the acceleration vector of the object, given in unit vector notation.
- **\(\vec{F}_1\) and \(\vec{F}_2\)**: Represent two out of the three forces acting on the object, given in unit vector notation (\(\hat{\imath}\) and \(\hat{\jmath}\) denote the x and y components, respectively).
- **Mass of the object**: 1.20 kg.
**Steps to Solve:**
1. **Apply Newton's Second Law**:
\[
\sum \vec{F} = m \vec{a}
\]
where \(m\) is the mass of the object, and \(\sum \vec{F}\) is the vector sum of all forces acting on the object.
2. **Calculate \(\sum \vec{F}\)**:
- Using the formula, \(m = 1.20 \, \text{kg}\) and \(\vec{a} = (4.80 \, \text{m/s}^2) \hat{\imath} + (5.00 \, \text{m/s}^2) \hat{\jmath}\).
\[
\sum \vec{F} = 1.20 \, \text{kg} \times \left[(4.80 \, \text{m/s
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