Question
What is the contact force between two blocks on a frictionless table when a horizontal force of 3.5 N is applied to the block with mass 2.5 kg, and the other block has a mass of 1.8 kg?
Answers
GPT-4o mini
To find the contact force between the two blocks on a frictionless table when a horizontal force is applied, we need to first determine the acceleration of the blocks.
**Step 1: Calculate the total mass.**
The total mass (\(m_{\text{total}}\)) is the sum of the masses of the two blocks.
\[
m_{\text{total}} = m_1 + m_2 = 2.5\, \text{kg} + 1.8\, \text{kg} = 4.3\, \text{kg}
\]
**Step 2: Calculate the acceleration of the system.**
Using Newton's second law (\(F = ma\)), we can find the acceleration (\(a\)) of the entire system when the horizontal force of \(3.5\, \text{N}\) is applied.
\[
a = \frac{F}{m_{\text{total}}} = \frac{3.5\, \text{N}}{4.3\, \text{kg}} \approx 0.814\, \text{m/s}^2
\]
**Step 3: Calculate the contact force.**
The contact force acts on the second block (mass \(m_2 = 1.8\, \text{kg}\)) due to the acceleration of the entire system. The force needed to accelerate the second block at this acceleration can be calculated as
\[
F_{\text{contact}} = m_2 \cdot a = 1.8\, \text{kg} \cdot 0.814\, \text{m/s}^2 \approx 1.465\, \text{N}
\]
**Final Result:**
Thus, the contact force between the two blocks is approximately \(1.47\, \text{N}\).
**Step 1: Calculate the total mass.**
The total mass (\(m_{\text{total}}\)) is the sum of the masses of the two blocks.
\[
m_{\text{total}} = m_1 + m_2 = 2.5\, \text{kg} + 1.8\, \text{kg} = 4.3\, \text{kg}
\]
**Step 2: Calculate the acceleration of the system.**
Using Newton's second law (\(F = ma\)), we can find the acceleration (\(a\)) of the entire system when the horizontal force of \(3.5\, \text{N}\) is applied.
\[
a = \frac{F}{m_{\text{total}}} = \frac{3.5\, \text{N}}{4.3\, \text{kg}} \approx 0.814\, \text{m/s}^2
\]
**Step 3: Calculate the contact force.**
The contact force acts on the second block (mass \(m_2 = 1.8\, \text{kg}\)) due to the acceleration of the entire system. The force needed to accelerate the second block at this acceleration can be calculated as
\[
F_{\text{contact}} = m_2 \cdot a = 1.8\, \text{kg} \cdot 0.814\, \text{m/s}^2 \approx 1.465\, \text{N}
\]
**Final Result:**
Thus, the contact force between the two blocks is approximately \(1.47\, \text{N}\).