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. 4 n ← 2.0 kg → 8 n; 6 n ← 0.50 kg → 4 n; 4 n ← 0.25 kg ← 4 n; 16.0 kg → 6 n → 2 n; 2 n ← 0.50 kg → 8 n; 6 n ← 0.25 kg → 2 n

Explanation:

Step1: Recall Newton's Second Law

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

Step2: Calculate for Object 1 (2.0 kg, 4N ←, 8N →)

Net force: \( F_{net1} = 8 - 4 = 4 \, \text{N (right)} \)
Acceleration: \( a_1 = \frac{4}{2.0} = 2 \, \text{m/s}^2 \)

Step3: Calculate for Object 2 (0.50 kg, 6N ←, 4N →)

Net force: \( F_{net2} = 4 - 6 = -2 \, \text{N (left)} \)
Acceleration: \( a_2 = \frac{-2}{0.50} = -4 \, \text{m/s}^2 \)

Step4: Calculate for Object 3 (0.25 kg, 4N ←, 4N ←)

Net force: \( F_{net3} = -4 - 4 = -8 \, \text{N (left)} \)
Acceleration: \( a_3 = \frac{-8}{0.25} = -32 \, \text{m/s}^2 \)

Step5: Calculate for Object 4 (16.0 kg, 6N →, 2N →? Wait, diagram: 16.0 kg, 6N →, 2N →? Wait, original: "16.0 kg →6N" and "→2N"? Wait, maybe typo, assuming 6N → and 2N ←? Wait, no, the diagram: "16.0 kg →6N" and "→2N"? Wait, maybe 6N right and 2N left? Wait, the user's diagram: "16.0 kg →6N" and "→2N" – maybe 6N right, 2N left. So net force: \( F_{net4} = 6 - 2 = 4 \, \text{N (right)} \)

Acceleration: \( a_4 = \frac{4}{16.0} = 0.25 \, \text{m/s}^2 \)

Wait, maybe I misread. Let's recheck. The fourth diagram: "16.0 kg →6N" and "→2N" – maybe both right? No, that would be 8N right. Wait, original: "16.0 kg →6N" and "→2N" – maybe 6N right, 2N left. Let's assume: 6N →, 2N ←. So \( F_{net} = 6 - 2 = 4 \, \text{N} \), \( a = 4/16 = 0.25 \, \text{m/s}^2 \).

Step6: Calculate for Object 5 (0.50 kg, 2N ←, 8N →)

Net force: \( F_{net5} = 8 - 2 = 6 \, \text{N (right)} \)
Acceleration: \( a_5 = \frac{6}{0.50} = 12 \, \text{m/s}^2 \)

Step7: Calculate for Object 6 (0.25 kg, 6N ←, 2N →)

Net force: \( F_{net6} = 2 - 6 = -4 \, \text{N (left)} \)
Acceleration: \( a_6 = \frac{-4}{0.25} = -16 \, \text{m/s}^2 \)

Wait, maybe I misread Object 4. Let's re-express all objects properly (assuming the diagrams are:

1. 2.0 kg: 4N ←, 8N →
2. 0.50 kg: 6N ←, 4N →
3. 0.25 kg: 4N ←, 4N ← (both left)
4. 16.0 kg: 6N →, 2N ← (so 6-2=4N right)
5. 0.50 kg: 2N ←, 8N → (so 8-2=6N right)
6. 0.25 kg: 6N ←, 2N → (so 2-6=-4N left)

Now list accelerations (sorted from most negative to most positive):

- Object 3: \( -32 \, \text{m/s}^2 \) (most negative, rank 1)
- Object 6: \( -16 \, \text{m/s}^2 \) (rank 2)
- Object 2: \( -4 \, \text{m/s}^2 \) (rank 3)
- Object 4: \( 0.25 \, \text{m/s}^2 \) (rank 4)
- Object 1: \( 2 \, \text{m/s}^2 \) (rank 5)
- Object 5: \( 12 \, \text{m/s}^2 \) (most positive, rank 6)

Wait, but let's recheck Object 4. If Object 4 is 16.0 kg with 6N right and 2N right (both right), then net force is 8N right, acceleration \( 8/16 = 0.5 \, \text{m/s}^2 \). But the original diagram: "16.0 kg →6N" and "→2N" – maybe both right. Then \( F_{net4} = 6 + 2 = 8 \, \text{N} \), \( a_4 = 8/16 = 0.5 \, \text{m/s}^2 \). But the key is to calculate each:

Let's redo all with correct net forces:

1. 2.0 kg: 8N →, 4N ←. \( F_{net} = 8 - 4 = 4 \, \text{N} \). \( a = 4/2 = 2 \, \text{m/s}^2 \) (→)
2. 0.50 kg: 4N →, 6N ←. \( F_{net} = 4 - 6 = -2 \, \text{N} \). \( a = -2 / 0.5 = -4 \, \text{m/s}^2 \) (←)
3. 0.25 kg: 4N ←, 4N ←. \( F_{net} = -4 -4 = -8 \, \text{N} \). \( a = -8 / 0.25 = -32 \, \text{m/s}^2 \) (←)
4. 16.0 kg: 6N →, 2N →? No, maybe 6N →, 2N ←. So \( F_{net} = 6 - 2 = 4 \, \text{N} \). \( a = 4 / 16 = 0.25 \, \text{m/s}^2 \) (→)
5. 0.50 kg: 8N →, 2N ←. \( F_{net} = 8 - 2 = 6 \, \text{N} \). \( a = 6 / 0.5 = 12 \, \text{m/s}^2 \) (→)
6. 0.25 kg: 2N →, 6N ←. \( F_{net} = 2 - 6 = -4 \, \text{N} \). \( a = -4 / 0.25 = -16 \, \text{m/s}^2 \) (←)

Now sort accelerations from most negative (smallest) to most positive (largest):

- \( a_3 = -32 \) (most negative, rank 1)
- \( a_6 = -16 \) (rank 2)
- \( a_2 = -4 \) (rank 3)
- \( a_4 = 0.25 \) (rank 4)
- \( a_1 = 2 \) (rank 5)
- \( a_5 = 12 \) (most positive, rank 6)

So the ranking (1=most negative, 6=most positive) is:
Object 3 (0.25 kg, 4N←,4N←): 1
Object 6 (0.25 kg, 2N→,6N←): 2
Object 2 (0.50 kg, 4N→,6N←): 3
Object 4 (16.0 kg, 6N→,2N←): 4
Object 1 (2.0 kg, 8N→,4N←): 5
Object 5 (0.50 kg, 8N→,2N←): 6

But since the problem is to rank by tapping, the key is calculating each acceleration. The accelerations (with sign) are:
-32, -16, -4, 0.25, 2, 12. So from most negative (1) to most positive (6):

1. Object 3 (0.25 kg, two left forces)
2. Object 6 (0.25 kg, 2 right, 6 left)
3. Object 2 (0.50 kg, 4 right, 6 left)
4. Object 4 (16.0 kg, 6 right, 2 left)
5. Object 1 (2.0 kg, 8 right, 4 left)
6. Object 5 (0.50 kg, 8 right, 2 left)

Answer:

The ranking (1 for most negative, 6 for most positive) is:

1. Object with 0.25 kg, 4N ←, 4N ← (acceleration -32 m/s²)
2. Object with 0.25 kg, 2N →, 6N ← (acceleration -16 m/s²)
3. Object with 0.50 kg, 4N →, 6N ← (acceleration -4 m/s²)
4. Object with 16.0 kg, 6N →, 2N ← (acceleration 0.25 m/s²)
5. Object with 2.0 kg, 8N →, 4N ← (acceleration 2 m/s²)
6. Object with 0.50 kg, 8N →, 2N ← (acceleration 12 m/s²)

(Note: The exact diagram labels may need adjustment, but the process is calculating net force and acceleration using \( a = F_{net}/m \) with sign based on direction.)