Understanding the Force of Friction Between a Block and the Floor

When faced with the scenario of a 2 kg block placed on a floor with a coefficient of friction of 0.4, and a horizontal force of 7 N applied to it, the key question is: what is the force of friction between the block and the floor? This article will explore this problem, offering a detailed explanation with mathematical calculations and relevant physics principles.

Step-by-Step Analysis

Step 1: Calculate the Normal Force

The normal force, which acts on the block, is equal to the weight of the block. The weight of the block can be calculated using the formula:

$$ F_N m cdot g $$

Where m 2 kg and the acceleration due to gravity, g 9.81 m/s2.

F_N  2 text{kg} cdot 9.81 text{m/s}^2  19.62 text{N}

Step 2: Calculate the Maximum Static Friction Force

The maximum static friction force can be calculated using the formula:

$$ F_{text{friction max}} mu cdot F_N $$

Substitute the known values:

$$ F_{text{friction max}} 0.4 cdot 19.62 text{N} 7.848 text{N} $$

Step 3: Compare Applied Force and Maximum Friction Force

Now, compare the applied force of 7 N with the maximum static friction force of 7.848 N:

Applied force: 7 N Maximum static friction force: 7.848 N

Since the applied force is less than the maximum static friction force, the block will not move. The friction force will be equal to the applied force, preventing motion.

Conclusion: The force of friction between the block and the floor is equal to the applied force, which is 7 N.

Additional Insight: Self-Adjusting Friction

It's important to note that friction is self-adjusting. The maximum static friction force is:

$$ F_{text{friction max}} mu cdot m cdot g 0.4 cdot 2 cdot 10 8 text{N} $$

However, this self-adjusting nature means that the actual friction force will be restricted to 3 N if the applied force is only 7 N. The block will only start moving when the applied force exceeds the maximum friction force.

Diagram and Equations for Clarity

Let's draw a diagram to visualize this scenario. The block experiences a normal force (R) from the floor, which balances the weight of the block. Hence, the normal reaction is equal to the weight of the block (2 kg x 9.81 m/s2).

Applying the concept of forces, we have:

$$ F_{text{applied}} 7 text{N} $$

The friction force in this case would be:

$$ F_{text{friction}} 3 text{N} $$

This is because the applied force (7 N) is less than the maximum friction force (8 N), and the friction force is self-adjusting to prevent motion.

Conclusion

This detailed analysis shows that the force of friction between the block and the floor is 7 N when a 7 N force is applied. It also highlights the principle of self-adjusting friction, where the friction force varies based on the applied force.