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. \( 0.6 \, \text{m/s}^2 \)
• 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

We use Newton's second law to find the acceleration: \[ F_{\text{net}} = ma \] Where \( F_{\text{net}} \) is the net force, \( m \) is the mass, and \( a \) is the acceleration. The applied force is 10 N, and the frictional force \( F_{\text{friction}} \) is given by: \[ F_{\text{friction}} = \mu N \] Where: - \( \mu = 0.2 \) is the coefficient of friction, - \( N = mg = 5 \times 9.8 = 49 \, \text{N} \) is the normal force (since the block is on a horizontal surface). Now, calculate the frictional force: \[ F_{\text{friction}} = 0.2 \times 49 = 9.8 \, \text{N} \] The net force acting on the block is: \[ F_{\text{net}} = 10 \, \text{N} - 9.8 \, \text{N} = 0.2 \, \text{N} \] Now apply Newton's second law: \[ a = \frac{F_{\text{net}}}{m} = \frac{0.2}{5} = 0.6 \, \text{m/s}^2 \] Thus, the acceleration of the block is \( 0.6 \, \text{m/s}^2 \).