a student, who weighs 600n, is standing on a bathroom scale and riding an elevator that is moving downwards with a speed that is decreasing at a rate of 2.8m/s². how many forces are acting on the student? what is the students mass? what is the net force on the student? which force does the scale read? what value does the scale read? (in newtons)

Question

Accepted Answer

### Sub - question 1: How many forces are acting on the student?
# Brief Explanations:
When a student is in an elevator, the forces acting on the student are gravity (acting downwards) and the normal force from the scale (acting upwards). So there are 2 forces.
# Answer:
2

### Sub - question 2: What is the student’s mass?
# Explanation:
## Step1: Recall the formula for weight
The weight $ W$ of an object is given by $ W = mg$, where $ g = 9.8\ m/s^{2}$ (acceleration due to gravity) and $ m$ is the mass. We know $ W=600\ N$.
## Step2: Solve for mass $ m$
Rearranging the formula $ m=\frac{W}{g}$. Substituting $ W = 600\ N$ and $ g=9.8\ m/s^{2}$, we get $ m=\frac{600}{9.8}\approx61.22\ kg$.
# Answer:
$ 61.22$ kg

### Sub - question 3: What is the net force on the student?
# Explanation:
## Step1: Determine the acceleration
The elevator is moving downwards with a decreasing speed, so the acceleration $ a$ is upwards. The magnitude of acceleration $ a = 2.8\ m/s^{2}$.
## Step2: Use Newton's second law $ F_{net}=ma$
We know $ m\approx61.22\ kg$ and $ a = 2.8\ m/s^{2}$. So $ F_{net}=61.22\ kg\times2.8\ m/s^{2}\approx171.42\ N$. Since the acceleration is upwards, the net force is upwards.
# Answer:
$ 171.42$ N

### Sub - question 4: Which force does the scale read?
# Brief Explanations:
The scale reads the normal force $ F_{N}$ that it exerts on the student.
# Answer:
Normal Force

### Sub - question 5: What value does the scale read? (in Newtons)
# Explanation:
## Step1: Analyze the forces on the student
The forces on the student are weight $ W = mg$ (downwards) and normal force $ F_{N}$ (upwards). The net force $ F_{net}=F_{N}-W$ (since $ F_{N}$ is up and $ W$ is down, and net force is up as acceleration is up).
## Step2: Rearrange to solve for $ F_{N}$
From $ F_{net}=F_{N}-W$, we get $ F_{N}=F_{net}+W$. We know $ F_{net}\approx171.42\ N$ and $ W = 600\ N$. So $ F_{N}=171.42\ N + 600\ N=771.42\ N$.
# Answer:
$ 771.42$ N

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QUESTION IMAGE

Question

Explanation:

Sub - question 1: How many forces are acting on the student?

Step1: Recall the formula for weight

Step2: Solve for mass \( m\)

Step1: Determine the acceleration

Step2: Use Newton's second law \( F_{net}=ma\)

Answer:

Sub - question 2: What is the student’s mass?