Struggling with this problem:
A student stands on a bathroom scale in an elevator at rest on the 64th floor of a building. The scale reads 848 N. As the elevator starts to move up, the scale reading increases to 968 N, then changes back to 848 N. What was the acceleration of the elevator?
add acceleration a to the g of gravity.
mg = 9.81m = 848, so m = 86.44 kg
86.44(g+a) = 968
a = 968/86.44 - 9.81 = 1.388 m/s^2
or, m(g+a)/mg = 968/848
a = 1.388
848 = m g
so m = 848/g
968 = m g + m a
m a = 968 - m g = 968 - 848 = 120
a = 120/m = (120 / 848 ) g
now just put in whatever g is on your planet
Well, it seems like the elevator is really playing mind games with that poor student on the bathroom scale. But don't worry, I'm here to help!
To solve this problem, we can use Newton's second law, which states that the Net Force is equal to the mass of an object multiplied by its acceleration (F=ma). In this case, the object is the student, and the net force is the difference between the scale readings.
Firstly, let's find the mass of the student. We know that weight is equal to mass times the acceleration due to gravity (W=mg). Since the scale reads 848 N, we can set that equal to the student's weight (mg). Assuming the acceleration due to gravity is 9.8 m/s^2, we can solve for the mass.
848 N = m * 9.8 m/s^2
Now, let's find the acceleration of the elevator. The difference in scale readings, 968 N and 848 N, represents the net force experienced by the student. Multiplying this net force by the student's mass will give us the acceleration.
Net Force = (968 N - 848 N) = 120 N
a = Net Force / mass
Plug in the values and calculate the acceleration. And voila! You'll have your answer.
Just remember, in the end, it's all about solving problems, not creating them.
To find the acceleration of the elevator, we can 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.
The student's weight, which is the force they exert on the scale, is equal to the product of their mass and the acceleration due to gravity. In this case, since the student is not moving vertically and is at rest, the weight is equal to the normal force exerted by the scale. We can represent this with the equation:
Weight = mass * acceleration due to gravity
Since the student's mass and the acceleration due to gravity are constant, the change in the scale reading can be attributed to the acceleration of the elevator.
1. Convert weight into mass:
Weight = mass * acceleration due to gravity
848 N = mass * 9.8 m/s^2
2. Solve for mass:
mass = 848 N / 9.8 m/s^2
Now, let's find the acceleration of the elevator during each phase of the motion:
1. From rest to the scale reading of 968 N:
3. Calculate the net force acting on the student:
Net force = Scale reading - Weight
Net force = 968 N - (mass * 9.8 m/s^2)
2. From the scale reading of 848 N back to 848 N:
4. Calculate the net force acting on the student:
Net force = Scale reading - Weight
Net force = 848 N - (mass * 9.8 m/s^2)
By comparing the net forces in both phases, we can find the acceleration of the elevator.
To find the acceleration of the elevator, we need to consider the forces acting on the student when the scale reading changes.
Let's break down the forces acting on the student in each scenario:
1. When the elevator is at rest on the 64th floor (848 N):
- There are two forces acting on the student: the gravitational force (weight) pulling the student downward (mg) and the normal force exerted by the scale upward (N).
- Since the elevator is at rest, the net force is zero. Hence, the normal force (N) must balance the weight (mg), where m is the mass of the student and g is the acceleration due to gravity (9.8 m/s^2). Therefore, we have N = mg.
2. When the elevator starts to move up (968 N):
- The gravitational force pulling the student downward (mg) remains the same.
- In addition to the gravitational force, there is now an upward force (F_up) being exerted on the student due to the acceleration of the elevator.
- The net force is now the sum of the gravitational force (mg) and the upward force (F_up). So we have mg + F_up = 968 N.
3. When the scale reading changes back to 848 N:
- At this point, the elevator is moving upward at a constant speed. So the net force is again zero, and the normal force (N) balances the weight (mg). Hence, N = mg.
Now, let's use these equations to find the acceleration (a) of the elevator:
From equation 1 (N = mg), we have N = mg = 848 N.
From equation 2 (mg + F_up = 968 N), we know that N = mg = 848 N, so we can write the equation as 848 N + F_up = 968 N.
Since the elevator is moving upward, the upward force (F_up) is greater than the gravitational force (mg). So we can rewrite the equation as:
F_up = 968 N - 848 N = 120 N.
Since F_up = ma (Newton's Second Law), where m is the mass of the student and a is the acceleration of the elevator, we can now solve for a:
120 N = ma.
Dividing both sides of the equation by m:
a = 120 N / m.
To find the acceleration (a), we need to know the mass of the student. Unfortunately, that information is not given in the problem. Therefore, without knowing the mass of the student, we cannot determine the exact acceleration of the elevator.