Two block of masses 3 kg and 1 kg are kept in contact with each other on a frictionless horizontal surface. If a force of 10 N is applied on the larger block what is the acceleration of the system? What is the contact force between the two blocks?

Options

A. `2.5 m//s^(2), 2.5 N`
B. `5 m//s^(2), 5 N`
C. `5 m//s^(2), 2.5 N`
D. `2.5 m//s^(2), 5N`

Answer

To solve the problem step by step, we will analyze the forces acting on both blocks and apply Newton's second law of motion.

### Step 1: Identify the system and the forces acting on it
We have two blocks:
- Block A (3 kg)
- Block B (1 kg)

A force of 10 N is applied to Block A. Since the surface is frictionless, we can ignore any frictional forces.

### Step 2: Draw the free body diagram
For Block A (3 kg):
- A force of 10 N is applied to the right.
- A contact force \( F \) acts to the left due to Block B.

For Block B (1 kg):
- The contact force \( F \) acts to the right.

### Step 3: Write the equations of motion
For Block A (3 kg):
Using Newton's second law, we can write:
\[ 10 \, \text{N} - F = 3 \, \text{kg} \cdot a \quad \text{(Equation 1)} \]

For Block B (1 kg):
Using Newton's second law, we can write:
\[ F = 1 \, \text{kg} \cdot a \quad \text{(Equation 2)} \]

### Step 4: Solve the equations
From Equation 2, we can express the contact force \( F \) in terms of acceleration \( a \):
\[ F = 1a \]

Now, substitute \( F \) from Equation 2 into Equation 1:
\[ 10 - 1a = 3a \]

Combine like terms:
\[ 10 = 3a + 1a \]
\[ 10 = 4a \]

Now, solve for \( a \):
\[ a = \frac{10}{4} = 2.5 \, \text{m/s}^2 \]

### Step 5: Find the contact force
Now, substitute the value of \( a \) back into Equation 2 to find the contact force \( F \):
\[ F = 1 \cdot 2.5 = 2.5 \, \text{N} \]

### Final Answers
- The acceleration of the system is \( 2.5 \, \text{m/s}^2 \).
- The contact force between the two blocks is \( 2.5 \, \text{N} \).