Physics · Laws Of Motion · NEET
No. The friction force f is measured in newtons and changes with the applied force. The coefficient of friction mu is just a number (like 0.15 or 0.5) that describes the pair of surfaces. They are linked by f = mu*N only at the limit. Below the limit, static friction is smaller than mu_s N and simply matches whatever push you apply.
No. mu = f/N is a force divided by a force, so the newtons cancel. It is dimensionless (a plain number). This is exactly why the 2018/2026 NEET statement 'coefficient of sliding friction has the dimensions of length' is WRONG. Never write mu in N or metres.
Static friction grows to match your push and keeps the body still. But it cannot grow forever. Its highest possible value is (fs)max = mu_s N, called limiting friction. If your push exceeds this, static friction can no longer balance it and the body starts to slide. So limiting friction is the 'breaking point' of static friction.
No, and this is the most common NEET mistake. f = mu_s N is only the maximum static value (at the point of slipping). When the body is at rest and not about to move, static friction is 0 <= fs <= mu_s N and equals the applied force. Only kinetic (sliding) friction stays fixed at f_k = mu_k N while moving.
At rest, the tiny contact points of the two surfaces settle and interlock more strongly, so you need a bigger push to break them free (larger mu_s). Once sliding starts, the surfaces skim over the bumps and interlock less, so friction drops. That is why it is harder to START pushing a heavy box than to KEEP it moving: mu_k < mu_s.
No. mu depends only on the nature of the two surfaces in contact (how rough or smooth they are). It does not depend on mass, weight, or the area of contact. Mass changes N (and hence the friction force f = mu*N), but mu itself stays the same for a given pair of surfaces.
Which one of the following statements is incorrect?
A body of mass m is kept on a rough horizontal surface (coefficient of friction mu). A horizontal force is applied but the body does not move. The magnitude of the resultant F of the normal reaction and the frictional force satisfies:
There are two inclined surfaces of equal length L and the same inclination 45 degrees with the horizontal; one is rough and the other perfectly smooth. A given body takes 2 times as much time to slide down the rough surface as the smooth surface. The coefficient of kinetic friction (mu_k) between the body and the rough surface is close to:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Limiting friction is the largest value static friction can reach, just before the body starts to slide. Its value is (fs)max = mu_s N, where mu_s is the coefficient of static friction and N is the normal reaction.
mu = f/N, where f is the friction force and N is the normal reaction. For static friction the limiting case gives mu_s = (fs)max / N; for kinetic friction mu_k = f_k / N. mu is a pure number with no units.
Because it is a force divided by a force. Both f and N are measured in newtons, so when you divide them the units cancel, leaving just a number. A NEET favourite trap is claiming it has the dimension of length, which is false.
The static coefficient is larger: mu_s > mu_k. It takes a bigger force to start a body moving (overcome limiting friction) than to keep it moving, because once sliding begins the surfaces interlock less.
For dry surfaces in the NEET model, no. The coefficient of friction and the friction force depend on the nature of the surfaces and the normal reaction N, not on the apparent area of contact.
