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?

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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.