What will happen to a person's weight when he is in a moving elevator at a constant speed? A. Increase B. Decrease C. Weight will not change D. May increase or decrease
C
When a person is in an elevator, their 'weight' can sometimes refer to their actual gravitational weight (the force of gravity acting on their mass) or their apparent weight (the normal force exerted by the elevator floor on them). The question asks what happens to a person's weight, implying the apparent weight, which is what a scale would measure.
Two main vertical forces act on the person inside the elevator:
Newton's Second Law of Motion states that the net force acting on an object is equal to the product of its mass and acceleration ($F_{net} = ma$). In the vertical direction, the net force is the difference between the upward normal force ($N$) and the downward gravitational force ($mg$).
Let's consider the upward direction as positive.
The net force equation is: $F_{net} = N - mg$
According to Newton's Second Law: $N - mg = ma$
The question specifies that the elevator is moving at a constant speed. Constant speed implies that the acceleration ($a$) of the elevator (and thus the person inside) is zero ($a = 0$). This is true regardless of whether the elevator is moving upwards or downwards, as long as the speed is not changing.
Substituting $a = 0$ into the equation from Newton's Second Law:
$N - mg = m \times 0$
$N - mg = 0$
$N = mg$
The normal force ($N$), which represents the apparent weight, is equal to the actual gravitational weight ($mg$) when the elevator is moving at a constant speed. This means the apparent weight does not change from what it would be if the person were standing on stationary ground.
Therefore, a person's weight (apparent weight) will not change when the elevator is moving at a constant speed, either upwards or downwards.
| Elevator Motion | Acceleration ($a$) | Net Force Equation ($N - mg = ma$) | Normal Force ($N$) / Apparent Weight |
|---|---|---|---|
| Stationary | $a=0$ | $N - mg = 0$ | $N = mg$ (No change) |
| Moving Up, Constant Speed | $a=0$ | $N - mg = 0$ | $N = mg$ (No change) |
| Moving Down, Constant Speed | $a=0$ | $N - mg = 0$ | $N = mg$ (No change) |
| Accelerating Upward | $a > 0$ (upward) | $N - mg = ma$ | $N = mg + ma$ (Increases) |
| Accelerating Downward | $a < 0$ (upward) or $a > 0$ (downward) | $N - mg = m(-|a|)$ or $mg - N = m|a|$ | $N = mg - m|a|$ (Decreases) |
Based on this analysis, when the elevator moves at a constant speed, the apparent weight is equal to the actual weight, meaning the weight will not change.
Let's quickly review the key points about weight in an elevator at constant speed:
The concept of apparent weight change is related to being in a non-inertial reference frame (a frame that is accelerating). An elevator accelerating upwards or downwards is a non-inertial frame. However, an elevator moving at a constant velocity (which includes constant speed in a straight line, and being stationary) is an inertial reference frame. In an inertial frame, the observed forces and Newton's laws hold true without needing to introduce 'fictitious' forces. When the elevator moves at a constant speed, you are essentially in an inertial frame relative to the ground, and your weight feels normal because the net force is zero when you are not accelerating relative to that frame.
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