Question

rank the objects according to acceleration; use 1 for the most negative and 8 for the most positive. directions: tap each diagram in order to toggle though th possible acceleration rankings. once each object has a unique ranking, a check answer button will appear. 6 n ⬅️ 0.50 kg ➡️ 6 n; 4.0 kg ➡️ 4 n, 6 n ➡️ 4.0 kg; 4 n ⬅️ 0.25 kg, 4 n ⬅️ 0.25 kg; 6 n ⬅️ 0.50 kg ➡️ 4 n; 8 n ⬅️ 4.0 kg ➡️ 4 n; 8 n ⬅️ 2.0 kg ➡️ 2 n; 2 n ⬅️ 0.50 kg ➡️ 8 n; 2 n ⬅️ 0.25 kg ➡️ 4 n

Explanation:

Step1: Recall Newton's Second Law

Newton's second law is \( F_{net} = ma \), so \( a=\frac{F_{net}}{m} \). We need to calculate the net force (\( F_{net} \)) for each object, then find acceleration (\( a \)) by dividing net force by mass (\( m \)). Let's define the positive direction as right and negative as left.

Step2: Calculate for each object:

Object 1 (Top - Left): \( m = 0.50 \, \text{kg} \), Forces: \( 6 \, \text{N (left)} \) and \( 6 \, \text{N (right)} \)

\( F_{net}= - 6 + 6 = 0 \, \text{N} \)
\( a_1=\frac{0}{0.50}=0 \, \text{m/s}^2 \)

Object 2 (Top - Right): \( m = 4.0 \, \text{kg} \), Forces: \( 6 \, \text{N (right)} \) and \( 4 \, \text{N (right)} \)

\( F_{net}=6 + 4 = 10 \, \text{N} \)
\( a_2=\frac{10}{4.0}=2.5 \, \text{m/s}^2 \)

Object 3 (Middle - Left): \( m = 0.25 \, \text{kg} \), Forces: \( 4 \, \text{N (left)} \) and \( 4 \, \text{N (left)} \)

\( F_{net}= - 4 - 4 = - 8 \, \text{N} \)
\( a_3=\frac{-8}{0.25}=-32 \, \text{m/s}^2 \)

Object 4 (Middle - Right): \( m = 0.50 \, \text{kg} \), Forces: \( 6 \, \text{N (left)} \) and \( 4 \, \text{N (right)} \)

\( F_{net}= - 6 + 4 = - 2 \, \text{N} \)
\( a_4=\frac{-2}{0.50}=-4 \, \text{m/s}^2 \)

Object 5 (Bottom - Left - Top): \( m = 4.0 \, \text{kg} \), Forces: \( 8 \, \text{N (left)} \) and \( 4 \, \text{N (right)} \)

\( F_{net}= - 8 + 4 = - 4 \, \text{N} \)
\( a_5=\frac{-4}{4.0}=-1 \, \text{m/s}^2 \)

Object 6 (Bottom - Left - Middle): \( m = 2.0 \, \text{kg} \), Forces: \( 8 \, \text{N (left)} \) and \( 2 \, \text{N (right)} \)

\( F_{net}= - 8 + 2 = - 6 \, \text{N} \)
\( a_6=\frac{-6}{2.0}=-3 \, \text{m/s}^2 \)

Object 7 (Bottom - Left - Bottom): \( m = 0.50 \, \text{kg} \), Forces: \( 2 \, \text{N (left)} \) and \( 8 \, \text{N (right)} \)

\( F_{net}= - 2 + 8 = 6 \, \text{N} \)
\( a_7=\frac{6}{0.50}=12 \, \text{m/s}^2 \)

Object 8 (Bottom - Right): \( m = 0.25 \, \text{kg} \), Forces: \( 2 \, \text{N (left)} \) and \( 4 \, \text{N (right)} \)

\( F_{net}= - 2 + 4 = 2 \, \text{N} \)
\( a_8=\frac{2}{0.25}=8 \, \text{m/s}^2 \)

Step3: Rank by acceleration (1 = most negative, 8 = most positive)

First, list accelerations:
\( a_3 = - 32 \) (most negative), \( a_4 = - 4 \), \( a_6 = - 3 \), \( a_5 = - 1 \), \( a_1 = 0 \), \( a_2 = 2.5 \), \( a_8 = 8 \), \( a_7 = 12 \) (most positive).

Now rank from most negative (1) to most positive (8):

1. Object 3 (\( a = - 32 \))
2. Object 4 (\( a = - 4 \))
3. Object 6 (\( a = - 3 \))
4. Object 5 (\( a = - 1 \))
5. Object 1 (\( a = 0 \))
6. Object 2 (\( a = 2.5 \))
7. Object 8 (\( a = 8 \))
8. Object 7 (\( a = 12 \))

Answer:

The ranking (from most negative (1) to most positive (8)) is:

1. Middle - Left (0.25 kg, two 4 N left)
2. Middle - Right (0.50 kg, 6 N left, 4 N right)
3. Bottom - Left - Middle (2.0 kg, 8 N left, 2 N right)
4. Bottom - Left - Top (4.0 kg, 8 N left, 4 N right)
5. Top - Left (0.50 kg, 6 N left, 6 N right)
6. Top - Right (4.0 kg, 6 N right, 4 N right)
7. Bottom - Right (0.25 kg, 2 N left, 4 N right)
8. Bottom - Left - Bottom (0.50 kg, 2 N left, 8 N right)

(Note: If the diagrams are labeled differently, adjust based on the visual, but the accelerations are calculated as above. The key is to assign 1 to the most negative acceleration (Object 3) and 8 to the most positive (Object 7).)