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

To find the acceleration of the block, we apply Newton's second law of motion. The net force acting on the block is the applied horizontal force minus the force of friction. The force of friction (\( f_{\text{friction}} \)) can be calculated using the formula:

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

where \( \mu \) is the coefficient of friction and \( N \) is the normal force. Since the block is on a horizontal surface, the normal force (\( N \)) is equal to the weight of the block (\( mg \)), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity. Thus:

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

Substitute the values into the friction formula:

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

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

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

According to Newton's second law:

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

where \( a \) is the acceleration. Solving for acceleration:

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

Answer 2

• Given: \[ m = 5\, \text{kg},\quad F_{\text{applied}} = 25\, \text{N},\quad \mu = 0.4,\quad g = 10\, \text{m/s}^2 \]
• Normal force: \( N = mg = 5 \times 10 = 50\, \text{N} \)
• Frictional force: \( f = \mu N = 0.4 \times 50 = 20\, \text{N} \)
• Net force: \( F_{\text{net}} = F_{\text{applied}} - f = 25 - 20 = 5\, \text{N} \)
• Acceleration: \[ a = \frac{F_{\text{net}}}{m} = \frac{5}{5} = 1\, \text{m/s}^2 \]