Question

rank the objects according to acceleration; use 1 for the most negative and 6 for the most positive.
directions: tap each diagram in order to toggle though the possible acceleration rankings. once each object has a unique ranking, a check answer button will appear.
0.25 kg →2 n → 6 n \t 2 n ← 0.25 kg → 6 n
8 n ← 0.50 kg 2 n ← \t 10.0 kg →2 n → 8 n
8 n ← 0.25 kg →2 n \t 4 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 for each object, then find acceleration. Let's define right as positive and left as negative.

Step2: Calculate for First Object (0.25 kg, 6N right, 2N right)

Net force \( F_{net1} = 6 + 2 = 8 \, N \) (both right, so positive).
Acceleration \( a_1 = \frac{8}{0.25} = 32 \, m/s^2 \).

Step3: Calculate for Second Object (0.25 kg, 6N right, 2N left)

Net force \( F_{net2} = 6 - 2 = 4 \, N \).
Acceleration \( a_2 = \frac{4}{0.25} = 16 \, m/s^2 \).

Step4: Calculate for Third Object (0.50 kg, 8N left, 2N left)

Net force \( F_{net3} = -8 - 2 = -10 \, N \) (both left, negative).
Acceleration \( a_3 = \frac{-10}{0.50} = -20 \, m/s^2 \).

Step5: Calculate for Fourth Object (10.0 kg, 8N right, 2N right)

Net force \( F_{net4} = 8 + 2 = 10 \, N \).
Acceleration \( a_4 = \frac{10}{10.0} = 1 \, m/s^2 \).

Step6: Calculate for Fifth Object (0.25 kg, 8N left, 2N right)

Net force \( F_{net5} = -8 + 2 = -6 \, N \).
Acceleration \( a_5 = \frac{-6}{0.25} = -24 \, m/s^2 \).

Step7: Calculate for Sixth Object (0.25 kg, 4N left, 4N left)

Net force \( F_{net6} = -4 - 4 = -8 \, N \).
Acceleration \( a_6 = \frac{-8}{0.25} = -32 \, m/s^2 \).

Step8: Rank by Acceleration (1 most negative, 6 most positive)

List accelerations:
\( a_6 = -32 \), \( a_5 = -24 \), \( a_3 = -20 \), \( a_4 = 1 \), \( a_2 = 16 \), \( a_1 = 32 \).

Ranking (1 to 6, 1 most negative, 6 most positive):
1: Sixth Object (\( a = -32 \))
2: Fifth Object (\( a = -24 \))
3: Third Object (\( a = -20 \))
4: Fourth Object (\( a = 1 \))
5: Second Object (\( a = 16 \))
6: First Object (\( a = 32 \))

Answer:

To rank, we use \( a = \frac{F_{net}}{m} \). After calculating each acceleration:

• Sixth Object (0.25 kg, 4N left + 4N left): \( a = -32 \, m/s^2 \) (rank 1)
• Fifth Object (0.25 kg, 8N left + 2N right): \( a = -24 \, m/s^2 \) (rank 2)
• Third Object (0.50 kg, 8N left + 2N left): \( a = -20 \, m/s^2 \) (rank 3)
• Fourth Object (10.0 kg, 8N right + 2N right): \( a = 1 \, m/s^2 \) (rank 4)
• Second Object (0.25 kg, 6N right + 2N left): \( a = 16 \, m/s^2 \) (rank 5)
• First Object (0.25 kg, 6N right + 2N right): \( a = 32 \, m/s^2 \) (rank 6)

(Note: The exact diagram labels might vary, but the ranking is based on acceleration values: most negative (1) to most positive (6).)