Question

A block of mass $ 5\, \text{kg} $ is placed on a rough horizontal surface. A horizontal force of $ 25\, \text{N} $ is applied. If the coefficient of friction is $ 0.4 $, what is the acceleration of the block? (Take $ g = 10\, \text{m/s}^2 $)

• A. \( 1\, \text{m/s}^2 \)
• B. \( 1\, \text{m/s}^2 \)
• C. \( 3\, \text{m/s}^2 \)
• D. \( 4\, \text{m/s}^2 \)

Hint:

Key Fact: Net acceleration = \( \frac{F - f}{m} \), where \( f = \mu mg \)

Answer 1

The Correct Option is B

Newton's second law is applied to determine the block's acceleration. The resultant force on the block is the applied horizontal force reduced by the frictional force. The frictional force (\( f_{\text{friction}} \)) is computed using the equation:

\( f_{\text{friction}} = \mu \times N \)

Here, \( \mu \) represents the coefficient of friction, and \( N \) is the normal force. Given the block is on a horizontal surface, the normal force (\( N \)) equals the block's weight (\( mg \)), with \( m \) being the mass and \( g \) the acceleration due to gravity. Consequently:

\( N = mg = 5 \, \text{kg} \times 10 \, \text{m/s}^2 = 50 \, \text{N} \)

Substituting these values into the friction equation yields:

\( f_{\text{friction}} = 0.4 \times 50 \, \text{N} = 20 \, \text{N} \)

The net force (\( F_{\text{net}} \)) acting on the block is the applied force less the frictional force:

\( F_{\text{net}} = 25 \, \text{N} - 20 \, \text{N} = 5 \, \text{N} \)

Based on Newton's second law:

\( F_{\text{net}} = ma \)

Where \( a \) signifies acceleration. Rearranging to solve for acceleration:

\( a = \frac{F_{\text{net}}}{m} = \frac{5 \, \text{N}}{5 \, \text{kg}} = 1 \, \text{m/s}^2 \)