Question

A man weighing 75 kg is standing in a lift. The weight of the man standing on a weighing machine kept in the lift when the lift is moving downwards freely under gravity is: 

• A. zero
• B. 75 kg
• C. 84.8 kg
• D. 65.2 kg

Answer 1

The Correct Option is A

The correct option is: (A) zero.

(i) The man's mass is m = 70 kg. The acceleration is a = 0. By applying Newton's second law of motion, the equation of motion becomes: R - mg = ma Here, ma represents the net force acting on the man. Since the lift moves at a uniform speed (a = 0), the equation simplifies to: R = mg = 70 × 10 = 700 N Hence, the reading on the weighing scale = 700 g = 700 / 10 = 70 kg.

(ii) The man's mass is m = 70 kg. The acceleration is a = 5 m/s² downward. Using Newton's second law of motion, we can rewrite the equation as: mg - R = ma Substituting the values: R = m(g - a) = 70(10 - 5) = 70 × 5 = 350 N Therefore, the reading on the weighing scale = 350 g = 350 / 10 = 35 kg.

(iii) The man's mass is m = 70 kg. The acceleration is a = 5 m/s² upward. Applying Newton's second law of motion gives us: R - mg = ma Substituting the values: R = m(g + a) = 70(10 + 5) = 70 × 15 = 1050 N Hence, the reading on the weighing scale = 1050 g = 1050 / 10 = 105 kg.

(iv) When the lift moves freely under gravity, a = g. By Newton's second law of motion, the equation becomes: mg - R = ma Substituting a = g: R = m(g - g) = 0 In this case, the reading on the weighing scale = 0 g = 0 kg. The man will experience a state of weightlessness.