A 750- N person steps on the weighing scale inside a moving elevator as it slows down to the ground floor. The person notices that his weight is 850 N while the elevator is decelerating How long will it take the elevator to reach the ground it it is moving at 25 km/h before the elevator started decelerating?
850 N up - 750 weight down = m a up
m a = 100 N
a = 100/m up
Vinitial = - 25 000m/ 3600 s = - 6.94 m/s
v = Vi + a t
assume v = 0 at ground with no crash
0 = -6.94 + (100/m) t
if g = 9.81m/s^2
then m = 750 N / 9.81 = 76.5 kg
so
(100/76.5) t = 6.94
t = 5.31 seconds
To find the time it takes for the elevator to reach the ground, we need to first determine the deceleration of the elevator.
Step 1: Convert the speed of the elevator to meters per second.
Given: Speed = 25 km/h
We know that 1 km/h is equal to 1000 m/3600 s (since there are 1000 meters in a kilometer and 3600 seconds in an hour).
So, converting the speed of the elevator to meters per second:
Speed = 25 km/h * 1000 m/3600 s
Speed = 25000 m/3600 s
Speed ≈ 6.94 m/s
Step 2: Calculate the net force acting on the person inside the elevator.
Given:
Weight of the person = 750 N
Weight when decelerating = 850 N
Since weight is the force exerted on an object, we can assume that the net force acting on the person inside the elevator is the difference in weights.
Net force = Weight when decelerating - Weight of the person
Net force = 850 N - 750 N
Net force = 100 N
Step 3: Use Newton's second law of motion to calculate the acceleration.
Newton's second law of motion states:
Net force = mass * acceleration
Given:
Net force = 100 N
Mass of the person = Weight of the person / acceleration due to gravity
Acceleration due to gravity is approximately 9.8 m/s².
Mass of the person = 750 N / 9.8 m/s²
Mass of the person ≈ 76.53 kg
Using Newton's second law of motion:
100 N = 76.53 kg * acceleration
Acceleration = 100 N / 76.53 kg
Acceleration ≈ 1.31 m/s²
Step 4: Use the kinematic equation to find the time.
The kinematic equation for motion with constant acceleration is:
v = u + at
Where:
v = final velocity (0 m/s, since the elevator comes to a stop at the ground)
u = initial velocity (6.94 m/s)
a = acceleration (-1.31 m/s²)
t = time (unknown)
We want to find t, so rearranging the equation:
t = (v - u) / a
t = (0 m/s - 6.94 m/s) / (-1.31 m/s²)
t ≈ 5.29 seconds
So, it will take approximately 5.29 seconds for the elevator to reach the ground.
To determine how long it will take for the elevator to reach the ground, we need to start by finding the acceleration of the elevator while decelerating.
We know that the weight of the person is equal to the normal force, which is the force exerted by the weighing scale. When the elevator is at rest or moving at a constant velocity, the weight and normal force will be the same. So, we have:
Weight of person = Normal force = 850 N
However, when the elevator is decelerating, there is an additional force acting on the person due to the deceleration. This force is given by the equation:
Force = mass × acceleration
The mass of the person can be calculated using the formula:
Weight = mass × gravity
where gravity is the acceleration due to gravity (approximately 9.8 m/s²). Rearranging the formula, we get:
mass = weight / gravity
mass = 850 N / 9.8 m/s²
mass ≈ 86.73 kg
Now, let's consider the forces acting on the person when the elevator is decelerating. We have:
Weight of person (850 N) - Normal force (750 N) = mass × acceleration
Substituting the values, we get:
850 N - 750 N = 86.73 kg × acceleration
100 N = 86.73 kg × acceleration
Rearranging the equation to solve for acceleration, we have:
acceleration = 100 N / 86.73 kg
acceleration ≈ 1.15 m/s²
Next, let's convert the initial velocity of the elevator from km/h to m/s. We know that 1 km/h is equal to 0.2778 m/s. Therefore, the initial velocity of the elevator is:
25 km/h × 0.2778 m/s = 6.945 m/s
To find the time it takes for the elevator to reach the ground, we can use the equation:
acceleration = (final velocity - initial velocity) / time
We need to solve for time, so rearrange the equation:
time = (final velocity - initial velocity) / acceleration
Substituting the values, we get:
time = (0 m/s - 6.945 m/s) / (-1.15 m/s²)
time ≈ 6.03 seconds
Therefore, it will take approximately 6.03 seconds for the elevator to reach the ground.