Question:easy

A man weighing 50 kg is in a lift moving down with an acceleration of $2.8 m s^{-2}$. The force exerted by the floor on him is:

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Lift moving down $\rightarrow$ Apparent weight decreases.
Updated On: Jun 10, 2026
  • 480 N
  • 240 N
  • 120 N
  • 350 N
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The Correct Option is D

Solution and Explanation

Step 1: Understand apparent weight.
A man of mass 50 kg stands in a lift moving downward and speeding up at $2.8$ m/s$^2$. The floor pushes up on him with a normal force, which we must find.

Step 2: Identify the two forces on the man.
Gravity pulls him down with weight $mg$. The floor pushes him up with normal force $N$. Since the lift accelerates downward, the net force points downward.

Step 3: Write Newton's second law.
Taking down as positive, the net downward force is $mg - N$, and this equals mass times acceleration: $mg - N = m a$.

Step 4: Rearrange for the floor force.
Solving gives $N = m(g - a)$. The floor pushes with less than full weight because the lift drops.

Step 5: Put in the numbers.
With $m = 50$ kg, $g = 9.8$ m/s$^2$, and $a = 2.8$ m/s$^2$: $N = 50 \times (9.8 - 2.8)$.

Step 6: Compute the answer.
$N = 50 \times 7 = 350$ N. \[ \boxed{350 \ \text{N}} \]
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