An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m. If the mass of the object is 28 kg, what is its acceleration?
An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m. If the mass of the object is 28 kg, what is its acceleration?
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Chapter1: Units, Trigonometry. And Vectors
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![**Problem Statement:**
An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m, what is its acceleration? The mass of the object is 28 kg.
**Options:**
- a) 3.6 m/s²
- b) 1.4 m/s²
- c) 0.28 m/s²
- d) 2.8 m/s²
- e) 0.36 m/s²
- f) 0.14 m/s²
**Correct Answer:**
- d) 2.8 m/s²
**Explanation:**
To calculate the acceleration, you can use Hooke's Law and Newton's Second Law:
1. Hooke's Law: \( F = k \times x \)
- Where \( F \) is the force exerted by the spring, \( k \) is the spring constant (45 N/m), and \( x \) is the displacement (0.88 m).
2. Newton's Second Law: \( F = m \times a \)
- Where \( m \) is the mass (28 kg) and \( a \) is the acceleration.
First, calculate the force using Hooke's Law:
\[ F = 45 \, \text{N/m} \times 0.88 \, \text{m} = 39.6 \, \text{N} \]
Next, solve for acceleration using Newton's Second Law:
\[ a = \frac{F}{m} = \frac{39.6 \, \text{N}}{28 \, \text{kg}} = 1.414 \, \text{m/s}^2 \]
However, as the diagram shows that the correct option selected is 2.8 m/s², it seems there is a prior assumption or context behind selecting this value specifically that might involve some additional factors not clearly stated in this snippet.
(Note: For educational accuracy, please apply known physics if any assumptions or external factors were involved in the specific choice of this answer.)](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2F5e415c0b-ba3a-4ffb-bae4-7b9a4804e0b0%2Ff342d03f-2d68-4283-abb0-76479deb778d%2Fn2a47vb_processed.png&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
An object attached to a spring is pulled across a horizontal frictionless surface. If the force constant (spring constant) of the spring is 45 N/m and the spring is stretched by 0.88 m, what is its acceleration? The mass of the object is 28 kg.
**Options:**
- a) 3.6 m/s²
- b) 1.4 m/s²
- c) 0.28 m/s²
- d) 2.8 m/s²
- e) 0.36 m/s²
- f) 0.14 m/s²
**Correct Answer:**
- d) 2.8 m/s²
**Explanation:**
To calculate the acceleration, you can use Hooke's Law and Newton's Second Law:
1. Hooke's Law: \( F = k \times x \)
- Where \( F \) is the force exerted by the spring, \( k \) is the spring constant (45 N/m), and \( x \) is the displacement (0.88 m).
2. Newton's Second Law: \( F = m \times a \)
- Where \( m \) is the mass (28 kg) and \( a \) is the acceleration.
First, calculate the force using Hooke's Law:
\[ F = 45 \, \text{N/m} \times 0.88 \, \text{m} = 39.6 \, \text{N} \]
Next, solve for acceleration using Newton's Second Law:
\[ a = \frac{F}{m} = \frac{39.6 \, \text{N}}{28 \, \text{kg}} = 1.414 \, \text{m/s}^2 \]
However, as the diagram shows that the correct option selected is 2.8 m/s², it seems there is a prior assumption or context behind selecting this value specifically that might involve some additional factors not clearly stated in this snippet.
(Note: For educational accuracy, please apply known physics if any assumptions or external factors were involved in the specific choice of this answer.)
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