Weighing yourself in the elevator

[test2morrow]

Need help:Weighing yourself in the elevator-acceleration problem

Homework Statement

If you stand on a spring scale in your bathroom at home, it reads 600N, which means your mass is 60kg. If instead, you stand on the scale while accelerating at 2 m/s^2 upward in an elevator, what would the scale read?

a. 120N
b. 480N
c. 600N
d. 720N

Homework Equations

W=mg

The Attempt at a Solution

So I asked my professor this question and he came up with some complicated answer that I didn't really understand. My own logic was that W=mg, W=60kg * (2m/s^2+9.8m/s^2), W=720N which is the correct answer. However, this is not the same steps the professor used which was something like F=N-W=ma and he did some subtraction work and somehow got 720N.

So can somebody explain to me the correct way to solve this problem?

 

[LowlyPion]

Homework Statement

If you stand on a spring scale in your bathroom at home, it reads 600N, which means your mass is 60kg. If instead, you stand on the scale while accelerating at 2 m/s^2 upward in an elevator, what would the scale read?

a. 120N
b. 480N
c. 600N
d. 720N

Homework Equations

W=mg

The Attempt at a Solution

So I asked my professor this question and he came up with some complicated answer that I didn't really understand. My own logic was that W=mg, W=60kg * (2m/s^2+9.8m/s^2), W=720N which is the correct answer. However, this is not the same steps the professor used which was something like F=N-W=ma and he did some subtraction work and somehow got 720N.

So can somebody explain to me the correct way to solve this problem?

That works too of course.

The 60 kg student is accelerating upward at 2 m/s². The Normal force of the student less the gravitational weight must yield an m*a of the same acceleration as the floor.
N is what the spring will register and rearranging that normal force = m*a + W = m*(g + a)

Your way works fine as well.

 

[nlightner]

As a scientist, it is important to approach problems with a clear and logical thought process. In this case, the problem is asking for the weight on a scale while accelerating in an elevator. The key concept to remember is that weight is a force and is equal to the mass of an object multiplied by the acceleration due to gravity (W=mg).

In this scenario, the weight on the scale is equal to the normal force acting on the person's feet. This normal force is equal to the person's weight when standing on a stationary scale. However, in an accelerating elevator, there is an additional force acting on the person's feet due to the acceleration. This force is equal to the person's mass multiplied by the acceleration of the elevator (F=ma).

So, to find the correct weight on the scale, we need to add the force due to acceleration to the person's weight. Using the equation F=ma, we can calculate this force by multiplying the person's mass (60kg) by the acceleration of the elevator (2m/s^2), resulting in a force of 120N.

Now, to find the weight on the scale, we add this force to the person's weight (mg). This gives us a total weight of 720N (600N from the person's weight + 120N from the force due to acceleration).

In conclusion, the correct way to solve this problem is to first calculate the force due to acceleration (F=ma) and then add it to the person's weight (mg). This will give you the correct weight on the scale, which in this case is 720N. It is important to use the correct equations and concepts when solving problems in science, as this ensures accurate and logical results.