Question

A man weighing \( W \) newtons entered a lift which moves with an acceleration of \( a \, \text{m/s}^2 \). When the lift is moving downward, the force exerted by the man on the floor of the lift is:

• A. \( W \left( 1 - \frac{a}{g} \right) \)
• B. \( W \left( 1 + \frac{a}{g} \right) \)
• C. \( W \left( 1 - \frac{g}{a} \right) \)
• D. \( W \left( 1 + \frac{g}{a} \right) \)

Hint:

When a lift accelerates downward, apparent weight decreases by \( ma \). Use \( N = m(g - a) \) for downward motion.

Answer 1

The Correct Option is A

Step 1: Define actual and apparent weight.
The actual weight of a man is the gravitational force: \[ W = mg \] The apparent weight is the normal reaction \( N \) by the floor of the lift.
Step 2: Apply Newton’s second law.
If the lift is accelerating downward with acceleration \( a \), the net force acting on the man in the vertical direction is: \[ mg - N = ma \Rightarrow N = m(g - a) \]
Step 3: Express normal force in terms of weight.
Since \( W = mg \), then: \[ m = \frac{W}{g} \Rightarrow N = \frac{W}{g}(g - a) = W\left(1 - \frac{a}{g}\right) \]
Step 4: Conclude the result.
So, the force exerted by the man on the floor of the lift (apparent weight) is: \[ \boxed{W \left(1 - \frac{a}{g} \right)} \]