007 (part 1 of 2)
An elevator accelerates upward at 1.2 m/s2. The acceleration of gravity is 9.8 m/s2 . What is the upward force exerted by the
floor of the elevator on a(n) 92 kg passenger? Answer in units of N
008 (part 2 of 2)
If the same elevator accelerates downwards with an acceleration of 1.2 m/s2, what is the upward force exerted by the elevator floor on the passenger?
Answer in units of N
ma= mg - N
N=mg -ma
791.2
To find the upward force exerted by the floor of the elevator on the passenger, you need to calculate the net force acting on the passenger in each scenario. The net force can be found using Newton's second law of motion:
Net force = mass × acceleration
For Part 1:
The upward acceleration of the elevator is given as 1.2 m/s^2 and the acceleration due to gravity is 9.8 m/s^2. Since the elevator is moving upward, the force of gravity acts downward on the passenger. To find the net force, we subtract the force due to gravity from the force due to the elevator's acceleration:
Net force = (mass × upward acceleration) - (mass × gravity)
Plugging in the given values:
Mass = 92 kg
Upward acceleration = 1.2 m/s^2
Gravity = 9.8 m/s^2
Net force = (92 kg × 1.2 m/s^2) - (92 kg × 9.8 m/s^2)
Simplifying the equation:
Net force = (110.4 N) - (901.6 N)
Net force = -791.2 N
Since the net force is negative, it indicates that the force is exerted downward on the passenger. The magnitude of the upward force exerted by the floor of the elevator on the passenger is 791.2 N.
For Part 2:
The elevator is now moving downward with an acceleration of 1.2 m/s^2. The force due to gravity still acts downward on the passenger, but the direction of the acceleration is opposite. Therefore, the net force will be the sum of the force due to gravity and the force due to the elevator's acceleration:
Net force = (mass × downward acceleration) + (mass × gravity)
Plugging in the given values:
Mass = 92 kg
Downward acceleration = 1.2 m/s^2
Gravity = 9.8 m/s^2
Net force = (92 kg × -1.2 m/s^2) + (92 kg × 9.8 m/s^2)
Simplifying the equation:
Net force = (-110.4 N) + (901.6 N)
Net force = 791.2 N
Since the net force is positive, it indicates that the force is exerted upward on the passenger. The magnitude of the upward force exerted by the elevator floor on the passenger is 791.2 N.
To determine the upward force exerted by the elevator floor on the passenger, we need to apply Newton's second law of motion, which states that the force (F) is equal to the mass (m) multiplied by the acceleration (a). In both parts of the question, the mass of the passenger remains the same at 92 kg, while the direction of acceleration changes.
For part 1:
Given:
Mass of the passenger (m) = 92 kg
Acceleration upwards (a) = 1.2 m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
To find the upward force in this scenario, we subtract the effect of gravity from the net upward acceleration experienced by the passenger.
Net upward acceleration = acceleration upwards - acceleration due to gravity
Net upward acceleration = 1.2 m/s^2 - 9.8 m/s^2 = -8.6 m/s^2
Now, we calculate the upward force using Newton's second law:
F = m * a
F = 92 kg * (-8.6 m/s^2)
F = -795.2 kg * m/s^2
Note that the negative sign indicates that the force is directed downwards. However, since the question asks for the magnitude of the force, we ignore the sign.
The upward force exerted by the floor of the elevator on the passenger is approximately 795.2 N.
For part 2:
Given:
Mass of the passenger (m) = 92 kg
Acceleration downwards (a) = 1.2 m/s^2
Acceleration due to gravity (g) = 9.8 m/s^2
Using the same logic as part 1, we need to find the net upward acceleration experienced by the passenger.
Net upward acceleration = acceleration upwards - acceleration due to gravity
Net upward acceleration = -1.2 m/s^2 - 9.8 m/s^2 = -11.0 m/s^2
Now we calculate the upward force using Newton's second law:
F = m * a
F = 92 kg * (-11.0 m/s^2)
F = -1012 kg * m/s^2
Again, the negative sign indicates the force is directed downwards, but we ignore the sign since we are looking for the magnitude.
The upward force exerted by the elevator floor on the passenger is approximately 1012 N.