A small block (m = 2 kg) is being pushed along (parallel to) a rough surface by a force with magnitude of 4.0 N. The acceleration is 1.2 m/s². If the force then increased to 5.0 N, determine the new acceleration. A. 2.1 m/s2 B. 2.3 m/s² C. 1.9 m/s² D. 1.7 m/s2 E. 3.2 m/s²
A small block (m = 2 kg) is being pushed along (parallel to) a rough surface by a force with magnitude of 4.0 N. The acceleration is 1.2 m/s². If the force then increased to 5.0 N, determine the new acceleration. A. 2.1 m/s2 B. 2.3 m/s² C. 1.9 m/s² D. 1.7 m/s2 E. 3.2 m/s²
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![**Problem Statement:**
A small block (m = 2 kg) is being pushed along (parallel to) a rough surface by a force with a magnitude of 4.0 N. The acceleration is 1.2 m/s². If the force then increased to 5.0 N, determine the new acceleration.
**Options:**
- A. 2.1 m/s²
- B. 2.3 m/s²
- C. 1.9 m/s²
- D. 1.7 m/s²
- E. 3.2 m/s²
**Solution Explanation:**
To solve this problem, use Newton's second law: \( F = ma \).
When the force is 4.0 N, and the mass \( m = 2 \, \text{kg} \), the initial acceleration \( a_1 = 1.2 \, \text{m/s}^2 \). So, considering the resistive force \( F_r \) due to friction:
\[ 4.0 - F_r = 2 \times 1.2 \]
\[ 4.0 - F_r = 2.4 \]
\[ F_r = 4.0 - 2.4 = 1.6 \, \text{N} \]
Now, if the force increases to 5.0 N:
\[ 5.0 - 1.6 = 2a_2 \]
\[ 3.4 = 2a_2 \]
\[ a_2 = 1.7 \, \text{m/s}^2 \]
Therefore, the new acceleration is **1.7 m/s²** (Option D).](/v2/_next/image?url=https%3A%2F%2Fcontent.bartleby.com%2Fqna-images%2Fquestion%2Fcf9dafc6-bac8-41ec-a856-62d58445bc24%2F5c97f3ea-edcd-48f1-bbf6-b2dfd4223f30%2Fvponp8b_processed.jpeg&w=3840&q=75)
Transcribed Image Text:**Problem Statement:**
A small block (m = 2 kg) is being pushed along (parallel to) a rough surface by a force with a magnitude of 4.0 N. The acceleration is 1.2 m/s². If the force then increased to 5.0 N, determine the new acceleration.
**Options:**
- A. 2.1 m/s²
- B. 2.3 m/s²
- C. 1.9 m/s²
- D. 1.7 m/s²
- E. 3.2 m/s²
**Solution Explanation:**
To solve this problem, use Newton's second law: \( F = ma \).
When the force is 4.0 N, and the mass \( m = 2 \, \text{kg} \), the initial acceleration \( a_1 = 1.2 \, \text{m/s}^2 \). So, considering the resistive force \( F_r \) due to friction:
\[ 4.0 - F_r = 2 \times 1.2 \]
\[ 4.0 - F_r = 2.4 \]
\[ F_r = 4.0 - 2.4 = 1.6 \, \text{N} \]
Now, if the force increases to 5.0 N:
\[ 5.0 - 1.6 = 2a_2 \]
\[ 3.4 = 2a_2 \]
\[ a_2 = 1.7 \, \text{m/s}^2 \]
Therefore, the new acceleration is **1.7 m/s²** (Option D).
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