Physics · Laws Of Motion · NEET
When the lift accelerates upward, the floor must not only hold your weight mg but also give you an extra upward push to make you accelerate up with the lift. So the floor pushes harder: N = m(g + a). Your body feels this bigger push from the floor as extra weight. This only happens while the lift is speeding up (a > 0). Once it moves at steady speed, a = 0 and the heavy feeling goes away.
No. Constant speed means acceleration a = 0, even if the lift is moving fast. With a = 0, Newton's second law gives N - mg = 0, so N = mg. The reading is exactly your real weight. Apparent weight only changes while the lift is speeding up or slowing down, not while it cruises. This is why you feel normal in the middle of a long lift ride.
When the lift accelerates down with acceleration a, take up as positive: N - mg = -ma, so N = m(g - a). The floor pushes up with less force than mg, so you feel lighter and a weighing machine reads less. If a is small, you feel only a little lighter; if a grows toward g, N drops toward zero.
In free fall the lift and you both accelerate down at a = g (the cable is cut). Put a = g into N = m(g - a): N = m(g - g) = 0. The floor gives no push at all, so a weighing machine reads zero and you feel weightless. You still have real weight mg pulling you down, but nothing pushes back on your feet, so there is no sensation of weight.
Look at the direction of the lift's acceleration, not its velocity. If acceleration points up (lift starting to go up, or slowing while going down), use N = m(g + a). If acceleration points down (lift starting to go down, or slowing while going up), use N = m(g - a). A safe method: take up as positive, write N - mg = ma with the correct sign of a, and solve for N.
A person standing on the floor of an elevator drops a coin. The coin reaches the floor in time t₁ if the elevator is at rest, and in time t₂ if the elevator is moving uniformly. Then:
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Apparent weight is the normal reaction N the floor (or weighing machine) exerts on you, which equals your real weight mg only when acceleration is zero.
Accelerating up: N = m(g + a) (heavier). Accelerating down: N = m(g - a) (lighter). At rest or constant speed: N = mg (normal). Free fall a = g: N = 0 (weightless).
Yes. When the lift accelerates upward, N = m(g + a) > mg, so a weighing machine reads more than your real weight.
If the lift accelerates downward faster than g (a > g), then N = m(g - a) is negative, meaning the floor would have to pull you down — only possible if you are strapped in. In a normal falling lift the maximum is free fall, where N = 0.
No. Mass m is constant everywhere. Only the apparent weight (the reading N) changes because the normal reaction changes with the lift's acceleration.
NEET regularly tests apparent weight through weighing-machine readings, coin-drop timing, and effective g inside accelerating lifts — all solved quickly with N = m(g ± a).
