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 the possible acceleration rankings. once each object has a unique ranking, a check answer button will appear. 2 n ← 0.25 kg → 6 n; 6 n ← 0.50 kg → 4 n; 6 n ← 0.50 kg → 6 n; 0.25 kg → 6 n → 2 n; 6 n ← 0.25 kg → 2 n; 8 n ← 2.0 kg → 2 n; 6 n ← 4 n ← 2.0 kg; 2 n ← 0.25 kg → 4 n

Explanation:

Step1: Recall Newton's Second Law

Newton's Second Law states \( F_{net} = ma \), so \( a = \frac{F_{net}}{m} \). We'll calculate the net force (\( F_{net} \)) for each object (taking right as positive) and then find acceleration.

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

\( F_{net} = 6 - 2 = 4 \, N \) (right positive).
\( a = \frac{4}{0.25} = 16 \, m/s^2 \).

Step3: Calculate for Second Object (0.50 kg, 6N left, 4N right)

\( F_{net} = 4 - 6 = -2 \, N \) (left negative).
\( a = \frac{-2}{0.50} = -4 \, m/s^2 \).

Step4: Calculate for Third Object (0.50 kg, 6N left, 6N right)

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

Step5: Calculate for Fourth Object (0.25 kg, 2N right, 6N right? Wait, diagram: 0.25 kg, 2N right? Wait, original: "0.25 kg →2N" and "→6N"? Wait, recheck: Fourth diagram: 0.25 kg, right arrows: 2N and 6N? Wait, maybe typo, but assume: 0.25 kg, forces: 6N right, 2N right? Wait, no, original: "0.25 kg →2N" and "→6N"? Wait, maybe "0.25 kg, 2N right, 6N right" → \( F_{net} = 6 + 2 = 8 \, N \)? Wait, no, left/right: first object: left 2N, right 6N. Fourth: "0.25 kg →2N" and "→6N" – maybe both right? So \( F_{net} = 6 + 2 = 8 \, N \). Then \( a = \frac{8}{0.25} = 32 \, m/s^2 \). Wait, maybe I misread. Wait, original fourth diagram: "0.25 kg →2N" and "→6N" – so two right forces? Then net force 8N.

Wait, no, let's re-express each diagram:

1. Object 1: 0.25 kg, \( F_{left}=2N \), \( F_{right}=6N \) → \( F_{net}=6-2=4N \), \( a=4/0.25=16 \, m/s^2 \).

2. Object 2: 0.50 kg, \( F_{left}=6N \), \( F_{right}=4N \) → \( F_{net}=4-6=-2N \), \( a=-2/0.5=-4 \, m/s^2 \).

3. Object 3: 0.50 kg, \( F_{left}=6N \), \( F_{right}=6N \) → \( F_{net}=0 \), \( a=0 \).

4. Object 4: 0.25 kg, \( F_{right}=6N \), \( F_{right}=2N \) (both right) → \( F_{net}=6+2=8N \), \( a=8/0.25=32 \, m/s^2 \).

5. Object 5: 0.25 kg, \( F_{left}=6N \), \( F_{right}=2N \) → \( F_{net}=2-6=-4N \), \( a=-4/0.25=-16 \, m/s^2 \).

6. Object 6: 2.0 kg, \( F_{left}=8N \), \( F_{right}=2N \) → \( F_{net}=2-8=-6N \), \( a=-6/2.0=-3 \, m/s^2 \).

7. Object 7: 2.0 kg, \( F_{left}=6N \), \( F_{left}=4N \) (both left) → \( F_{net}=-(6+4)=-10N \)? Wait, no, diagram: "6N ← 2.0 kg 4N ←" → both left, so \( F_{net}=-6-4=-10N \), \( a=-10/2.0=-5 \, m/s^2 \). Wait, maybe I misread. Wait, original seventh diagram: "6N ← 2.0 kg 4N ←" → left forces: 6N and 4N, so net force \( -10N \), \( a=-10/2=-5 \, m/s^2 \).

8. Object 8: 0.25 kg, \( F_{left}=2N \), \( F_{right}=4N \) → \( F_{net}=4-2=2N \), \( a=2/0.25=8 \, m/s^2 \).

Now, list accelerations:

- Object 5: \( -16 \, m/s^2 \) (most negative, rank 1)
- Object 7: \( -5 \, m/s^2 \) (rank 2)
- Object 2: \( -4 \, m/s^2 \) (rank 3)
- Object 6: \( -3 \, m/s^2 \) (rank 4)
- Object 3: \( 0 \, m/s^2 \) (rank 5)
- Object 8: \( 8 \, m/s^2 \) (rank 6)
- Object 1: \( 16 \, m/s^2 \) (rank 7)
- Object 4: \( 32 \, m/s^2 \) (rank 8, most positive)

Wait, but let's recheck Object 5: 0.25 kg, left 6N, right 2N → \( F_{net}=2-6=-4N \)? Wait, no: 2N right, 6N left → \( F_{net}=2-6=-4N \), \( a=-4/0.25=-16 \, m/s^2 \). Yes, that's correct.

Object 7: 2.0 kg, left 8N? Wait, original: "8 N ← 2.0 kg →2N" → \( F_{net}=2-8=-6N \), \( a=-6/2=-3 \, m/s^2 \). Oh, I misread earlier! Object 6: "8 N ← 2.0 kg →2N" → \( F_{net}=2-8=-6N \), \( a=-6/2=-3 \, m/s^2 \).

Object 7: "6 N ← 2.0 kg 4 N ←" → left 6N and 4N, so \( F_{net}=-(6+4)=-10N \), \( a=-10/2=-5 \, m/s^2 \).

Object 5: "6 N ← 0.25 kg →2N" → \( F_{net}=2-6=-4N \), \( a=-4/0.25=-16 \, m/s^2 \).

Object 2: "6 N ← 0.50 kg →4N" → \( F_{net}=4-6=-2N \), \( a=-2/0.5=-4 \, m/s^2 \).

Now, correct accelerations:

1. Object 5 (0.25 kg, 6N left, 2N right): \( a = (2 - 6)/0.25 = -16 \, m/s^2 \) (rank 1)
2. Object 7 (2.0 kg, 6N left, 4N left): \( F_{net}= -10N \), \( a = -10/2 = -5 \, m/s^2 \) (rank 2)
3. Object 2 (0.50 kg, 6N left, 4N right): \( a = -4 \, m/s^2 \) (rank 3)
4. Object 6 (2.0 kg, 8N left, 2N right): \( a = -3 \, m/s^2 \) (rank 4)
5. Object 3 (0.50 kg, 6N left, 6N right): \( a = 0 \) (rank 5)
6. Object 8 (0.25 kg, 2N left, 4N right): \( F_{net}=4-2=2N \), \( a=2/0.25=8 \, m/s^2 \) (rank 6)
7. Object 1 (0.25 kg, 2N left, 6N right): \( F_{net}=6-2=4N \), \( a=16 \, m/s^2 \) (rank 7)
8. Object 4 (0.25 kg, 2N right, 6N right): \( F_{net}=8N \), \( a=32 \, m/s^2 \) (rank 8)

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

Object 5 (rank 1), Object 7 (rank 2), Object 2 (rank 3), Object 6 (rank 4), Object 3 (rank 5), Object 8 (rank 6), Object 1 (rank 7), Object 4 (rank 8).

Answer:

To rank the objects by acceleration (1 = most negative, 8 = most positive), we use \( a = \frac{F_{net}}{m} \) (right = positive):

1. Object 5 (0.25 kg, 6N left, 2N right): \( a = \frac{2 - 6}{0.25} = -16 \, m/s^2 \) (rank 1)
2. Object 7 (2.0 kg, 6N left, 4N left): \( a = \frac{-10}{2} = -5 \, m/s^2 \) (rank 2)
3. Object 2 (0.50 kg, 6N left, 4N right): \( a = \frac{4 - 6}{0.50} = -4 \, m/s^2 \) (rank 3)
4. Object 6 (2.0 kg, 8N left, 2N right): \( a = \frac{2 - 8}{2} = -3 \, m/s^2 \) (rank 4)
5. Object 3 (0.50 kg, 6N left, 6N right): \( a = 0 \, m/s^2 \) (rank 5)
6. Object 8 (0.25 kg, 2N left, 4N right): \( a = \frac{4 - 2}{0.25} = 8 \, m/s^2 \) (rank 6)
7. Object 1 (0.25 kg, 2N left, 6N right): \( a = \frac{6 - 2}{0.25} = 16 \, m/s^2 \) (rank 7)
8. Object 4 (0.25 kg, 2N right, 6N right): \( a = \frac{8}{0.25} = 32 \, m/s^2 \) (rank 8)

(Note: Diagrams may have typos, but the key is calculating \( F_{net} = F_{right} - F_{left} \) and \( a = F_{net}/m \), then ranking from most negative to most positive.)