Question

If a 5 kg mass is suspended by a spring balance in a lift which is acceleration downwards at 10 m/s². The reading of the balance-(g=10m/s²)

• A. More than 5 kg weight
• B. Is less than 5 kg weight
• C. Is equal to 5 kg weigth
• D. Zero

Answer 1

Answer: (D)

Zero

Answer 2

The forces acting on the 5 kg mass are the tension $T$ exerted by the spring balance upwards and the gravitational force $mg$ downwards. The lift is accelerating downwards with acceleration $a = 10$ m/s². Given $g = 10$ m/s². Applying Newton's second law of motion in the downward direction: $mg - T = ma$. The reading of the spring balance is the tension $T$. $T = mg - ma = m(g - a)$. Substituting the given values: $m = 5$ kg, $g = 10$ m/s², $a = 10$ m/s². $T = 5 \text{ kg} \times (10 \text{ m/s}^2 - 10 \text{ m/s}^2) = 5 \text{ kg} \times (0 \text{ m/s}^2) = 0 \text{ N}$. The reading of the balance is 0 N, which means it reads zero.