It takes the elevator in a skyscraper 4.5 seconds to reach its cruising speed of 11 m/s. A 70 kg passenger gets aboard on the ground floor.
A. What is the passenger's weight before the elevator starts moving?
B. What is the passenger's weight while the elevator is speeding up?
C. What is the passenger's weight after the elevator reaches its cruising speed?
Answer (3)
A and C are very simple. Since there is no acceleration in both cases, the weight of the passenger is just m g = ( 70 k g ) ( 9.8 s 2 m ) = 686 N (about 690 N or 150 lbs with correct sig-figs). For B, we must first find the acceleration of the elevator. Since v = v 0 + a t , we can write that 11 s m = 0 s m + a ( 4.5 s ) . Solving for a, we get that a ≈ 2.44 s 2 m . Plugging this into Newton's second law, we get that F = ma = ( 70 k g ) ( 2.44 s 2 m ) ≈ 171 N . When calculating the apparent weight, this gets added onto the 686 N from A and C, so the final apparent weight is about 857 N, or 860 N with the correct number of sig-figs.
tl;dr: A - 690 N B - 860 N C - 690 N
The question pertains to the calculation of a passenger's weight in different conditions of elevator movement. Let's address each part of the question.
A- What is the passenger's weight before the elevator starts moving?
Before the elevator starts moving, the passenger's weight is simply the force due to gravity acting on them. This can be calculated using the equation Weight = mass × gravity, where mass is 70kg, and we'll use the standard gravitational acceleration of 9.8m/s². Thus, the weight = 70kg × 9.8m/s² = 686 N (Newtons).
B- What is the passenger's weight while the elevator is speeding up?
While the elevator is accelerating, the apparent weight felt by the passenger changes. The rate of acceleration is found by dividing the change in speed by the time taken, which is (11m/s) / (4.5s) = 2.44m/s². Using the equation Fnet = mass × (gravity + acceleration), we get an apparent weight of 70kg × (9.8m/s² + 2.44m/s²) = 857.8 N, reflecting the increased force the passenger feels due to the elevator's acceleration.
C- What is the passenger's weight after the elevator reaches its cruising speed?
Once the elevator reaches its cruising speed and is moving at a constant velocity, the acceleration is 0. Therefore, the passenger's apparent weight returns to its actual weight due to gravity, which is 686 N, since there is no additional acceleration acting on it.
The passenger's weight before the elevator starts moving is 686 N. While the elevator speeds up, the apparent weight is approximately 857 N. After reaching cruising speed, the weight returns to 686 N.
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