Question

A man weighs 80 kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of 6 m/s\(^2\). What would be his weight in kg? (g = 10 m/s\(^2\))

• A. Zero
• B. 48 kg
• C. 120 kg
• D. 128 kg

Hint:

Apparent weight is the normal force exerted by the scale. Always be clear if the question asks for apparent weight in Newtons or apparent mass in kg (which is the apparent weight divided by \( g \)).

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks for the apparent weight of a man (in kg, which implies a mass reading on a scale) in a lift accelerating upwards.
Step 2: Key Formula or Approach:
Apparent weight in a lift accelerating upwards: \( N = m(g + a) \), where \( N \) is the normal force exerted by the scale.
If the scale reads "weight in kg", it is effectively measuring mass \( m' \) such that \( m'g = N \).
So, apparent mass \( m' = \frac{N}{g} = \frac{m(g+a)}{g} = m \left(1 + \frac{a}{g}\right) \).
Step 3: Detailed Explanation:
Given values:
Mass of man, \( m = 80 \text{ kg} \)
Acceleration of lift, \( a = 6 \text{ m/s}^2 \) (upwards)
Acceleration due to gravity, \( g = 10 \text{ m/s}^2 \)
Calculate the apparent mass (\( m' \)):
\[ m' = 80 \left(1 + \frac{6}{10}\right) \] \[ m' = 80 \left(1 + 0.6\right) \] \[ m' = 80 \times 1.6 \] \[ m' = 128 \text{ kg} \]
Step 4: Final Answer:
His weight (apparent mass) in kg would be 128 kg.