KCET · Physics · Laws of Motion
A man weighs \(80\) kg. He stands on a weighing scale in a lift which is moving upwards with a uniform acceleration of \(6 \text{ m/s}^2\). What would be his weight in kg? (\(g = 10 \text{ m/s}^2\))
- A Zero
- B \(48\) kg
- C \(120\) kg
- D \(128\) kg
Answer & Solution
Correct Answer
(D) \(128\) kg
Step-by-step Solution
Detailed explanation
Mass of the man, \(m = 80 \text{ kg}\)
Acceleration of the lift, \(a = 6 \text{ m/s}^2\) upwards
The weighing scale measures the normal reaction \(N\). The equation of motion for the man is:
\(N - mg = ma\)
\(N = m(g + a)\)
Substituting the given values:
\(N = 80(10 + 6) = 1280 \text{ N}\)
The reading of the weighing scale in kg is:
\(\dfrac{N}{g} = \dfrac{1280}{10} = 128 \text{ kg}\)
Answer: \(128\) kg
Acceleration of the lift, \(a = 6 \text{ m/s}^2\) upwards
The weighing scale measures the normal reaction \(N\). The equation of motion for the man is:
\(N - mg = ma\)
\(N = m(g + a)\)
Substituting the given values:
\(N = 80(10 + 6) = 1280 \text{ N}\)
The reading of the weighing scale in kg is:
\(\dfrac{N}{g} = \dfrac{1280}{10} = 128 \text{ kg}\)
Answer: \(128\) kg
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