Final answer:
The weight a scale shows in an elevator varies depending on acceleration. For deceleration over 25m at a starting speed of 36km/h, it shows approximately 590N; over 12.5m, it's greater than 590N; moving upwards it shows around 390N; and at a constant speed, it reflects actual weight.
Step-by-step explanation:
When weighing a person in an elevator, the scale reading can vary depending on the acceleration of the elevator. Since weight is a force, it is calculated by multiplying mass by the acceleration due to gravity. However, if the elevator is accelerating, this must be considered in the weight calculation.
When braking from the 2nd to the first floor with a deceleration over 25m:
The initial speed is 36km/h, which is 10m/s when converted to m/s. Using the equation v^2 = u^2 + 2as, where v is the final velocity (0 m/s), u is the initial velocity (10 m/s), a is the acceleration, and s is the distance (25 m), we can solve for 'a'. After finding the acceleration, we use F=ma to calculate the force exerted by the scale, hence the reading. In this case, the reading would be approximately 590N as stated.
When braking with a deceleration over 12.5m:
The process is similar to above, but the distance is now 12.5m. The deceleration would be greater, and thus the scale would show a heavier weight than 590N.
When moving in the opposite direction:
If the elevator moves upward and decelerates over the 25m distance, the opposite direction of acceleration will cause a different force reading on the scale - around 390N as stated - since the person would effectively feel lighter.
At a constant speed:
When the elevator moves at a constant speed, the reading on the scale will be equal to the actual weight of the person because there is no additional acceleration, just the standard acceleration due to gravity.