Question

A block of mass 10 kg placed on the rough horizontal surface having a coefficient of friction µ = 0.5, if a horizontal force of 100 N acting on it then the acceleration of the block will be :

• A. 10 m/s²
• B. 5 m/s²
• C. 15 m/s²
• D. 0.5 m/s²

Answer 1

The Correct Option is B

To find the acceleration of the block, we first need to calculate the net force acting on the block. The forces acting on the block are:

1. The applied force (F_{\text{applied}} = 100 \, \text{N}).
2. The frictional force acting against the motion.

The frictional force (F_{\text{friction}}) can be calculated using the formula:

F_{\text{friction}} = \mu \cdot N,

where \mu is the coefficient of friction, and N is the normal force. For a block on a horizontal surface, the normal force (N) is equal to the weight of the block (N = mg). Thus:

N = m \cdot g = 10 \, \text{kg} \cdot 9.8 \, \text{m/s}^2 = 98 \, \text{N}

Therefore, the frictional force is:

F_{\text{friction}} = 0.5 \cdot 98 = 49 \, \text{N}

Now, the net force (F_{\text{net}}) acting on the block can be calculated as follows:

F_{\text{net}} = F_{\text{applied}} - F_{\text{friction}}

F_{\text{net}} = 100 \, \text{N} - 49 \, \text{N} = 51 \, \text{N}

Using Newton's second law of motion, the acceleration (a) of the block can be calculated by:

F_{\text{net}} = m \cdot a \Rightarrow a = \frac{F_{\text{net}}}{m}

Substituting the known values, we get:

a = \frac{51 \, \text{N}}{10 \, \text{kg}} = 5.1 \, \text{m/s}^2

To simplify, this is approximately 5 m/s². Hence, the correct answer is:

5 m/s²