A Force 18 N pulls a 40 kg mass on a horizontal floor at 0.3 ms$^{-2}$. Find the coefficient of friction.

1. Analyze the Question

We are given:

• Applied Force ($F$) = 18 N
• Mass ($m$) = 40 kg
• Acceleration ($a$) = 0.3 ms$^{-2}$
• We need to find the coefficient of friction ($\mu$).

2. Explain the Concept

When a force pulls an object on a surface, the net force acting on the object is the difference between the applied force and the frictional force ($F_f$). According to Newton's Second Law of Motion:

The frictional force is defined as $F_f = \mu R$, where $R$ is the normal reaction. On a horizontal floor, $R = mg$ (where $g$ is acceleration due to gravity, approximately 10 ms$^{-2}$).

3. Show the Solution

Step 1: Calculate the net force ($F_{net}$)

Step 2: Calculate the frictional force ($F_f$) Using the equation $F - F_f = F_{net}$:

Step 3: Calculate the coefficient of friction ($\mu$) We know $F_f = \mu mg$. Using $g = 10 \, \text{ms}^{-2}$:

4. Identify the Answer

Comparing our result to the options provided, the correct value is 0.015. This matches option B.

5. Exam Tip

Always remember that the "Applied Force" is not the same as the "Net Force." The applied force must first overcome friction before it can cause acceleration. If you ever get a negative frictional force, double-check your subtraction!