Henry, whose mass is 95  kg , stands on a bathroom scale in an elevator. The scale reads 830  N  for the first 4.0  s  after the elevator starts moving, then 930  N  for the next 4.0  s. What is the elevator's speed 8.0  s  after starting?

To determine the elevator's speed 8.0 seconds after it starts moving, we need to break the problem into two parts: the first 4 seconds and the next 4 seconds. We'll use Newton's second law of motion (\( F = ma \)) and kinematic equations to analyze Henry's motion.

### Known Data:
- Mass of Henry, \( m = 95 \, \text{kg} \)
- Scale reading for the first 4 seconds, \( F_1 = 830 \, \text{N} \)
- Scale reading for the next 4 seconds, \( F_2 = 930 \, \text{N} \)
- Gravitational acceleration, \( g = 9.8 \, \text{m/s}^2 \)

### 1. Analyze Forces:
The reading on the scale represents the normal force exerted by the scale, which is the combination of Henry's weight and the effect of acceleration due to the elevator's motion.

#### (a) First 4 seconds:
The scale reads \( F_1 = 830 \, \text{N} \). The difference between this force and Henry’s actual weight provides the net force due to acceleration:

- Henry's weight: 
\[
W = mg = 95 \times 9.8 = 931 \, \text{N}
\]

- The net force on Henry during the first 4 seconds:
\[
F_{\text{net1}} = F_1 - W = 830 - 931 = -101 \, \text{N}
\]

This net force causes an acceleration, which can be found using Newton's second law:
\[
F_{\text{net1}} = ma_1 \implies a_1 = \frac{F_{\text{net1}}}{m} = \frac{-101}{95} = -1.06 \, \text{m/s}^2
\]

So, the acceleration during the first 4 seconds is \( a_1 = -1.06 \, \text{m/s}^2 \) (downward).

#### (b) Next 4 seconds:
The scale now reads \( F_2 = 930 \, \text{N} \).

- The net force on Henry during the next 4 seconds:
\[
F_{\text{net2}} = F_2 - W = 930 - 931 = -1 \, \text{N}
\]

This net force causes an acceleration:
\[
a_2 = \frac{F_{\text{net2}}}{m} = \frac{-1}{95} = -0.01 \, \text{m/s}^2
\]

So, the acceleration during the next 4 seconds is \( a_2 = -0.01 \, \text{m/s}^2 \) (almost negligible).

### 2. Kinematic Analysis:

Now, we can find the velocity of the elevator after the 8 seconds.

#### (a) Velocity after the first 4 seconds:
We assume the elevator starts from rest (\( v_0 = 0 \)).

Using the kinematic equation:
\[
v_1 = v_0 + a_1 t_1 = 0 + (-1.06)(4) = -4.24 \, \text{m/s}
\]

#### (b) Velocity after the next 4 seconds:
The velocity after the next 4 seconds can be found using:
\[
v_2 = v_1 + a_2 t_2 = -4.24 + (-0.01)(4) = -4.24 - 0.04 = -4.28 \, \text{m/s}
\]

### Final Answer:
The elevator’s speed 8 seconds after it starts moving is approximately \( 4.28 \, \text{m/s} \) downward.

Do you want more details or have any questions?

---

Here are 5 related questions you might find interesting:
1. How would the elevator's motion change if the scale read a higher value during the second interval?
2. What would happen to the readings if the elevator moved upwards?
3. How does Newton’s second law help us understand acceleration in non-freefall situations?
4. If Henry’s mass were different, how would that affect the scale readings?
5. What if the elevator moved at a constant speed instead of accelerating?

**Tip:** Always start by analyzing forces and accelerations separately in multi-step motion problems.

Henry, whose mass is 95 kg, stands on a bathroom scale in an elevator. The scale reads 830 N for the first 4.0 s after the elevator starts moving, then 930 N for the next 4.0 s. What is the elevator's speed 8.0 s after starting?

This problem demonstrates how to calculate the speed of an elevator after 8 seconds using Newton's second law and kinematic equations. Henry stands on a scale that reads 830 N and 930 N over two intervals. Suitable for students in Grades 9-12.

Math Problem Statement

Solution

Ask a new question for Free

By Image

Math Problem Analysis

Mathematical Concepts

Formulas

Theorems

Suitable Grade Level

Related Recommendation