Question

The weight of a man in a lift moving upwards with an acceleration 'a' is 620 N. When the lift moves downwards with the same acceleration, his weight is found to be 340 N. The real weight of the man is

• A. 620 N
• B. 680 N
• C. 380 N
• D. 480 N

Hint:

When an object moves up and down with identical acceleration rates, the pseudo-forces cancel out completely if you take the average. The real weight is simply the exact arithmetic mean of the two apparent weights: $\frac{620 + 340}{2} = 480\ \text{N}$!

Answer 1

The Correct Option is D

Step 1: Write the apparent weights.
Going up: $m(g+a)=620$. Going down: $m(g-a)=340$.

Step 2: Add the two equations.
The acceleration $a$ cancels: $m(g+a)+m(g-a)=2mg$. So $620+340=2mg$.

Step 3: Solve for the real weight.
$960=2mg$, so $mg=480$ N. \[ \boxed{480\text{ N}} \]