Friction
ICSE · Class 11 · Physics
Flashcards for Friction — ICSE Class 11 Physics. Quick Q&A cards covering key concepts, definitions, and formulas.
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What is friction?
Answer
Friction is an opposing force that comes into play when one body actually moves or even tries to move over the surface of another body. It acts tangential to the contact surface and opposite to the di…
Why does a block at rest sometimes still experience friction?
Answer
When a force is applied on a body at rest, static friction adjusts itself to oppose the applied force. It can be zero if no force is applied, but it becomes nonzero as soon as a force tries to move th…
State the formula for limiting static friction.
Answer
Limiting static friction: f_s^{max} = μ_s R Meaning: f_s^{max} is the maximum static frictional force, μ_s is the coefficient of static friction, and R is the normal reaction. Units: f_s^{max} in newt…
A 20 kg block is just about to move on a rough horizontal surface when a force of 98 N is applied. Find μ_s and angle of friction.
Answer
Given: m = 20 kg, F = 98 N, g = 9.8 m/s^2 Since the block is just about to move, F = f_s^{max}. Normal reaction: R = mg = 20 kg × 9.8 m/s^2 = 196 N Step 1: μ_s = f_s^{max} / R = 98 N / 196 N = 0.5 Ste…
State the law of limiting friction.
Answer
The magnitude of limiting static friction depends on the nature and roughness of the surfaces in contact and not on the size or area of the surfaces. It is directly proportional to the normal reaction…
Why is friction independent of contact area in everyday solids?
Answer
The actual contact area is much smaller than the apparent contact area, roughly in the ratio 1:10^4. As the apparent area changes, the microscopic points of actual contact adjust, so friction remains …
What does the modern theory say about the origin of friction?
Answer
According to modern theory, friction arises from strong atomic or molecular adhesive forces at the points of actual contact. These points form cold welds, and extra force is needed to break them for m…
A 2 kg block is on a floor. μ_s = 0.4. A horizontal force of 2.5 N is applied. Find the frictional force.
Answer
Given: m = 2 kg, μ_s = 0.4, applied force F = 2.5 N, g = 9.8 m/s^2 Step 1: Normal reaction R = mg = 2 kg × 9.8 m/s^2 = 19.6 N Step 2: Maximum static friction f_s^{max} = μ_s R = 0.4 × 19.6 N = 7.84 N …
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