Spring Scale in a Moving Elevator
A 2.0 kg mass hangs from a calibrated spring scale inside an elevator. Starting from rest, a student records the scale reading as the elevator moves. The reading first rises to 23.5 N for 1.2 s, then settles at 19.6 N for 2.0 s, drops to 15.7 N for 1.5 s, and returns to 19.6 N. The student claims that because the reading is below the weight during the low interval, the elevator must be moving downward then, and that the smallest reading corresponds to the greatest downward speed. The scale is accurate to ±0.2 N; the time intervals are measured with a stopwatch to ±0.1 s. The student has no independent measurement of velocity or position. A second student argues that the scale reveals acceleration, not velocity, so the same reading pattern could occur while the elevator moves either up or down.
1. Calculate the acceleration of the mass during the 15.7 N interval, including its direction. Can the elevator’s velocity direction be determined from that interval alone? Explain.
Read the worked answer
Take upward as positive. The true weight is
mg = (2.0 kg)(9.8 m/s²) = 19.6 N.
During the low interval the scale exerts N = 15.7 N upward, so the net vertical force is
F_net = N − mg = 15.7 − 19.6 = −3.9 N.
Newton’s second law gives
a = F_net/m = −3.9/2.0 = −1.95 m/s²,
so the acceleration is about 1.95 m/s² downward.
This tells us the direction of acceleration, not velocity. A downward acceleration can occur while the elevator moves upward and slows, while it moves downward and speeds up, or at the instant it is momentarily at rest. Therefore the velocity direction cannot be determined from this interval alone.
2. Evaluate the two claims: (a) a reading below the weight implies the elevator is moving downward; (b) the smallest reading corresponds to the greatest downward speed. Use the full reading sequence to support your answer.
Read the worked answer
Both claims are incorrect.
(a) A reading below the true weight means downward acceleration, not downward velocity. The elevator could be moving upward while slowing down. The scale measures the normal force and therefore apparent weight; it does not measure velocity directly.
(b) The reading is constant at 15.7 N for 1.5 s, so the acceleration is constant during that interval. Speed changes continuously, so there is no single speed associated with the smallest reading.
Using upward as positive and starting from rest:
- First interval, N = 23.5 N: a = (23.5 − 19.6)/2.0 = +1.95 m/s². After 1.2 s, v = +2.34 m/s.
- Middle interval, N = 19.6 N: a = 0, so v stays +2.34 m/s for 2.0 s.
- Low interval, N = 15.7 N: a = −1.95 m/s² for 1.5 s. Then v changes from +2.34 m/s to +2.34 − (1.95)(1.5) = −0.585 m/s.
So the elevator is still moving upward at the start of the low-reading interval and is moving downward by the end of it. After the reading returns to 19.6 N, the elevator continues downward at constant speed. A below-weight reading can therefore occur while the elevator moves upward, and the lowest reading does not by itself mean the greatest downward speed.
3. The student wants to decide whether the elevator is moving up or down without looking outside. Is the spring scale plus the known start from rest sufficient? Describe a valid check and a limitation.
Read the worked answer
Yes, in principle, if the elevator truly starts from rest and the scale is read often enough. The scale gives acceleration through a = (N − mg)/m. Integrating a(t) from a known initial velocity of zero gives velocity, including its sign. The interval data above already give v = +2.34 m/s after the first interval and v = −0.585 m/s after the low interval.
A valid check is to record the scale reading at short time steps and numerically integrate the acceleration, or to use an accelerometer and integrate its output. The limitation is that this depends on the initial velocity actually being zero and on the measurements being frequent enough to catch changes. A motion sensor fixed inside the elevator would not settle the question, because it measures motion relative to the elevator, not the elevator’s velocity relative to the ground. An external reference, such as tracking position relative to the building, would give velocity directly.