Question

A 5 kg block is placed on a horizontal surface. A force of 10 N is applied to the block. The coefficient of friction between the block and the surface is 0.2. Find the acceleration of the block.

• A. \( 0.04 \, \text{m/s}^2 \)
• B. \( 1.0 \, \text{m/s}^2 \)
• C. \( 2.0 \, \text{m/s}^2 \)
• D. \( 1.5 \, \text{m/s}^2 \)

Hint:

The frictional force always opposes the motion of the object and reduces the effective applied force. Make sure to account for friction when calculating the net force.

Answer 1

The Correct Option is A

Given:

• Mass of the block: \( m = 5 \, \text{kg} \)
• Applied force: \( F_{\text{applied}} = 10 \, \text{N} \)
• Coefficient of friction: \( \mu = 0.2 \)
• Acceleration due to gravity: \( g = 9.8 \, \text{m/s}^2 \)

Step 1: Calculate the Normal Force

The normal force (\( N \)) is equal to the weight of the block: \[ N = m \cdot g = 5 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 49 \, \text{N} \]

Step 2: Calculate the Frictional Force

The frictional force (\( F_{\text{friction}} \)) is given by: \[ F_{\text{friction}} = \mu \cdot N = 0.2 \times 49 \, \text{N} = 9.8 \, \text{N} \]

Step 3: Calculate the Net Force

The net force (\( F_{\text{net}} \)) is the applied force minus the frictional force: \[ F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}} = 10 \, \text{N} - 9.8 \, \text{N} = 0.2 \, \text{N} \]

Step 4: Calculate the Acceleration

Using Newton’s Second Law (\( F = ma \)): \[ a = \frac{F_{\text{net}}}{m} = \frac{0.2 \, \text{N}}{5 \, \text{kg}} = 0.04 \, \text{m/s}^2 \]

✅ Final Answer:

The acceleration of the block is \( \boxed{0.04 \, \text{m/s}^2} \).