Apparent weight in a moving lift (elevator): A body that truly weighs 14 g (measured at rest) appears to weigh 13 g when measured by a spring balance in a moving lift. Determine the magnitude of the lift’s acceleration at that instant (state the closest standard value).

Difficulty: Medium

Correct Answer: 0.7 m/s^2

Explanation:


Introduction / Context:
Apparent weight changes in an accelerating frame such as a lift (elevator). A spring balance reads the normal reaction on the body, which differs from the true weight when the frame accelerates. This classic dynamics problem reinforces Newton’s second law in non-inertial scenarios treated from an inertial frame.


Given Data / Assumptions:

  • True weight corresponds to mass m such that W = m g; the balance reads 14 g at rest.
  • Apparent weight during motion is equivalent to 13 g on the same balance.
  • Take g ≈ 9.8 m/s^2 (ratios will cancel).
  • Motion is vertical; friction and air resistance neglected.


Concept / Approach:

When the lift accelerates downward with acceleration a, the apparent weight N satisfies N = m (g − a). When it accelerates upward, N = m (g + a). A reading lower than the true weight indicates downward acceleration (or upward acceleration with less likely opposing sign choice). Use the ratio of apparent to true weight to solve for a/g.


Step-by-Step Solution:

True weight W_true = m g corresponds to “14 g units” on the balance.Apparent weight W_app = m (g − a) corresponds to “13 g units”.Form the ratio: W_app / W_true = (g − a)/g = 13/14.Solve for a: g − a = (13/14) g → a = g (1 − 13/14) = g / 14.Compute: a ≈ 9.8 / 14 ≈ 0.7 m/s^2 (downward).


Verification / Alternative check:

If the lift were accelerating upward, apparent weight would exceed true weight (contrary to the observation). Thus the direction is downward with magnitude about 0.7 m/s^2, consistent with the sign logic.


Why Other Options Are Wrong:

0.5 and 1.0 m/s^2 do not satisfy the precise ratio 13/14; 0.01 m/s^2 is negligible; 1.5 m/s^2 is too large for the given readings.


Common Pitfalls:

Confusing mass units with force units; using W = m a incorrectly; forgetting that a reduced reading implies downward acceleration.


Final Answer:

0.7 m/s^2

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