A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 840 N. As the elevator moves up, the scale reading increases to 943 N, then decreases back to 840 N.
The acceleration of gravity is 9.8 m/s^2.
Find the acceleration of the elevator.
Answer in units m/s^2.
As the elevator approaches 74th floor the scale reading drops as low as 790 N.
What is the acceleration of the elevator?
Answer in units of m/s^2
To find the acceleration of the elevator, we need to use Newton's second law of motion, which states that the net force acting on an object is equal to the mass of the object multiplied by its acceleration.
1. First, let's find the mass of the student. The weight of an object is given by the formula W = mg, where W is the weight, m is the mass, and g is the acceleration due to gravity. Since the student's weight is given, we can rearrange the formula to solve for mass: m = W / g.
Given: weight (W) = 840 N, acceleration due to gravity (g) = 9.8 m/s^2.
m = 840 N / 9.8 m/s^2 ≈ 85.71 kg.
2. Let's consider the first scenario where the scale reading increases to 943 N. In this case, the scale reading is equal to the sum of the weight of the student and the force applied by the elevator (equivalent to the student's acceleration times the mass).
Let's assume the acceleration of the elevator is a.
943 N = (85.71 kg)(9.8 m/s^2) + (85.71 kg)(a).
Simplifying the equation:
943 N = 840 N + (85.71 kg)(a).
103 N = (85.71 kg)(a).
Therefore, the acceleration of the elevator in this scenario is a ≈ 1.20 m/s^2.
3. Let's now consider the second scenario where the scale reading drops to 790 N. Again, the scale reading is equal to the sum of the weight of the student and the force applied by the elevator.
790 N = (85.71 kg)(9.8 m/s^2) + (85.71 kg)(a).
Simplifying the equation:
790 N = 840 N + (85.71 kg)(a).
-50 N = (85.71 kg)(a).
Therefore, the acceleration of the elevator in this scenario is a ≈ -0.58 m/s^2. Note that the negative sign indicates that the elevator is decelerating (moving in the opposite direction of its acceleration).
So, the acceleration of the elevator as it approaches the 74th floor is approximately -0.58 m/s^2.
To find the acceleration of the elevator in the given scenarios, we'll use Newton's second law of motion, which states that the net force acting on an object is equal to its mass multiplied by its acceleration:
F_net = m * a
In this case, the object is the student standing on the scale inside the elevator. We'll assume the mass of the student to be m.
Let's break down the scenarios and calculate the acceleration of the elevator in each case:
1) The scale reads 840 N when the elevator is at rest on the 64th floor.
In this scenario, we have the following forces acting on the student:
- Weight (W) of the student, which is equal to m * g, where g is the acceleration due to gravity (9.8 m/s^2).
- Normal force (N) exerted by the scale on the student, which is equal to the force the scale reads (840 N).
Since the elevator is at rest (not accelerating), the net force on the student is zero. Therefore, we can equate the weight and the normal force:
N = W
Since we are given the weight of the student (840 N), we can substitute the formula for weight:
N = m * g
840 N = m * 9.8 m/s^2
Now, we can solve for the mass (m):
m = 840 N / 9.8 m/s^2
m ≈ 85.71 kg
Since the elevator is at rest, the acceleration of the elevator in this scenario is 0 m/s^2.
2) The scale reads 943 N and then decreases back to 840 N.
In this scenario, the elevator is accelerating upwards. As the elevator accelerates, the net force on the student increases.
Using the same equations as before, we can equate the weight and the normal force:
N = W
Since we are given the weight of the student at the maximum scale reading (943 N), we can substitute the formula for weight:
N = m * g
943 N = m * 9.8 m/s^2
Now, we can solve for the mass (m):
m = 943 N / 9.8 m/s^2
m ≈ 96.22 kg
Since the net force is greater than the weight, there must be an additional force acting on the student, which is the net force due to the acceleration of the elevator. Let's denote this net force as F_net.
F_net = N - W
F_net = 943 N - (m * g) (substituting the given values)
To find the acceleration (a), we can use Newton's second law:
F_net = m * a
Substituting the known values, we obtain:
943 N - (m * g) = m * a
Simplifying the equation:
943 N - (96.22 kg * 9.8 m/s^2) = 96.22 kg * a
a ≈ (943 N - 943.156 kg*m/s^2) / 96.22 kg
a ≈ -0.156 m/s^2 (approximately)
The negative sign indicates that the acceleration of the elevator is in the downward direction.
3) The scale reading drops to 790 N as the elevator approaches the 74th floor.
Using the same approach as before, the net force acting on the student is given by:
F_net = N - W
F_net = 790 N - (m * g)
To find the acceleration (a), we can use Newton's second law:
F_net = m * a
Substituting the known values, we obtain:
790 N - (m * g) = m * a
Simplifying the equation:
790 N - (96.22 kg * 9.8 m/s^2) = 96.22 kg * a
a ≈ (790 N - 943.156 kg*m/s^2) / 96.22 kg
a ≈ -1.588 m/s^2 (approximately)
Again, the negative sign indicates that the acceleration of the elevator is in the downward direction.
To summarize:
1) The acceleration of the elevator when the scale reads 840 N is 0 m/s^2.
2) The acceleration of the elevator when the scale reads 943 N and then decreases back to 840 N is approximately -0.156 m/s^2 (downward).
3) The acceleration of the elevator when the scale reading drops to 790 N as the elevator approaches the 74th floor is approximately -1.588 m/s^2 (downward).