2.
A box of mass m slides along a wooden floor. The contact area between the box and the
floor is A. Which of these statements is true about the frictional force?
a. It doubles if A doubles while m is kept the same
b. It doubles if m doubles while A is kept the same
c. It does not change if A doubles and m decreases by half
d. It does not change if both A and m doubles
A box of mass m slides along a wooden floor with contact area A, raising key questions about frictional force dependence on mass versus area. Kinetic friction f_k = μ_k mg confirms independence from A, as real contact occurs only at surface asperities.
Physics Formula
Normal force: N = mg
Therefore: f_k = μ_k mg (independent of contact area A)
Option Analysis
Doubling A while keeping m constant leaves N = mg unchanged, so f_k stays the same—not doubles. False.
Doubling m while keeping A constant doubles N to 2mg, so f_k doubles to 2μ_k mg. True.
Doubling A and halving m halves N to 0.5mg, so f_k halves—not unchanged. False.
Doubling both A and m doubles N to 2mg, so f_k doubles—not unchanged. False.
✅ Correct Answer: Option b
The frictional force acting on a box of mass m sliding on a wooden floor with contact area A follows Coulomb’s law of friction, f_k = μ_k N, where N = mg is the normal force.
Key Physics Principles
- Coulomb Friction Laws: Force proportional to normal reaction N = mg, independent of macroscopic area. Doubling mass doubles friction; area changes do not.
- Wooden Floor Context: Typical μ_k for wood-wood is 0.2-0.4, but principle holds universally for dry friction.
Common Misconceptions
- Larger A seems intuitive for more friction, but pressure P = mg/A balances it exactly.
- In exams like CSIR NET Physics, test area independence rigorously.
This resolves the MCQ: only mass doubling affects frictional force on the sliding box.
