It takes the elevator in a skyscraper 4.5s to reach its cruising speed of 11m/s . A 70kg passenger gets aboard on the ground floor.
What is the passenger's weight before the elevator starts moving?
What is the passenger's weight while the elevator is speeding up?
To find the passenger's weight before the elevator starts moving, we need to understand that weight is the force acting on an object due to gravity. The formula to calculate weight is:
Weight = mass × acceleration due to gravity
Now, the acceleration due to gravity is approximately 9.8 m/s². The mass of the passenger is given as 70 kg. So, the weight of the passenger can be calculated as:
Weight = 70 kg × 9.8 m/s²
Hence, the passenger's weight before the elevator starts moving is 686 Newtons.
Now, let's move on to the passenger's weight while the elevator is speeding up. When the elevator is accelerating, there are two forces acting on the passenger: the force of gravity (weight) and the force due to the elevator's acceleration. This additional force causes an increase in apparent weight or "weight" felt by the passenger.
Using Newton's second law of motion, we can calculate the net force acting on the passenger:
Net Force = mass × acceleration
Here, the mass of the passenger remains the same, 70 kg. The elevator's acceleration can be calculated using the formula:
Acceleration = change in velocity / time taken
The change in velocity is the difference between the cruising speed and initial speed, which is zero. So the acceleration of the elevator is:
Acceleration = (11 m/s - 0 m/s) / 4.5 s
Hence, the acceleration of the elevator is approximately 2.44 m/s².
Now, let's calculate the net force acting on the passenger:
Net Force = 70 kg × 2.44 m/s²
Therefore, the passenger's weight while the elevator is speeding up is 170.8 Newtons.