a basket of apples is pulled with a constant force. a friction force acts in a direction opposite to the motion. the basket starts at rest and increases its speed over time (accelerates). click on each action that will reduce the acceleration (rate at which the basket increases speed over time). options: a pull with less force, b pull with more force, c add an apple to the basket, d take an apple out of the basket, e smooth the surface of the table to decrease friction, f roughen the surface of the table to increase friction.

Question

Accepted Answer

To solve this, we use Newton's second law $ F_{net} = ma $ (net force equals mass times acceleration) and analyze how each action affects net force or friction (which opposes motion, reducing net force):

### Step 1: Recall Newton’s Second Law
Net force $ F_{net} = F_{pull} - F_{friction} $, and $ a = \frac{F_{net}}{m} $. To *reduce* acceleration, we need to reduce $ F_{net} $ (either by increasing friction, increasing mass, or decreasing the pulling force).

### Analyze Each Option:
- **Option A (pull with less force):** Decreasing $ F_{pull} $ reduces $ F_{net} $ (since $ F_{net} = F_{pull} - F_{friction} $). Lower $ F_{net} $ means lower acceleration ($ a = \frac{F_{net}}{m} $). This reduces acceleration.
- **Option B (pull with more force):** Increases $ F_{pull} $, so $ F_{net} $ increases. Acceleration would *increase*, not decrease. Eliminate.
- **Option C (add an apple to the basket):** Increases mass ($ m $). From $ a = \frac{F_{net}}{m} $, if $ F_{net} $ stays the same and $ m $ increases, $ a $ *decreases*. Wait, but let’s check other options. Wait, adding mass increases inertia, but does it affect net force? Wait, friction is $ F_{friction} = \mu N $, and $ N $ (normal force) equals weight ($ mg $). So adding mass increases $ N $, which increases friction. Wait, no—wait, the basket is being pulled horizontally. Normal force $ N = mg $ (vertical). So adding mass increases $ m $, so $ N $ increases, so friction $ F_{friction} = \mu N $ increases. Also, mass $ m $ increases. Let’s re-express $ F_{net} = F_{pull} - \mu mg $, so $ a = \frac{F_{pull} - \mu mg}{m} = \frac{F_{pull}}{m} - \mu g $. If $ m $ increases, $ \frac{F_{pull}}{m} $ decreases, and $ \mu g $ is constant. So $ a $ decreases. But wait, the question is which action *reduces* acceleration. But let’s check Option E and F.
- **Option D (take an apple out):** Decreases mass ($ m $). From $ a = \frac{F_{pull} - \mu mg}{m} = \frac{F_{pull}}{m} - \mu g $, decreasing $ m $ increases $ \frac{F_{pull}}{m} $, so $ a $ *increases*. Eliminate.
- **Option E (smooth the table):** Decreases friction ($ \mu $ decreases, since smoother surface has less friction). So $ F_{friction} = \mu mg $ decreases, so $ F_{net} = F_{pull} - F_{friction} $ increases. Acceleration *increases*. Eliminate.
- **Option F (roughen the table):** Increases friction ($ \mu $ increases). So $ F_{friction} $ increases, $ F_{net} $ decreases, so $ a $ decreases. Wait, but now we have two options (A, C, F?) Wait, no—let’s re-examine the problem statement: the basket is pulled with a *constant force*. Wait, the original scenario: “A basket of apples is pulled with a constant force. A friction force acts in a direction opposite to the motion. The basket starts at rest and increases its speed over time (accelerates).” We need to find which action *reduces* acceleration (i.e., makes it speed up more slowly, or even slow down, but since it’s accelerating initially, reducing acceleration means smaller positive acceleration).

Wait, let’s re-express $ a = \frac{F_{pull} - F_{friction}}{m} $.

- **Option A (pull with less force):** $ F_{pull} $ decreases → $ F_{net} $ decreases → $ a $ decreases.
- **Option C (add an apple):** $ m $ increases, and $ F_{friction} = \mu mg $ increases (since $ N = mg $ increases). So $ F_{net} = F_{pull} - \mu mg $. If $ m $ increases, both the numerator ($ F_{pull} - \mu mg $) and denominator ($ m $) change. Let’s take numbers: Suppose $ F_{pull} = 10 N $, $ \mu = 0.2 $, initial $ m = 2 kg $. Then initial $ F_{friction} = 0.2*2*9.8 = 3.92 N $, $ F_{net} = 10 - 3.92 = 6.08 N $, $ a = 6.08 / 2 = 3.04 m/s² $. Add an apple: $ m = 3 kg $, $ F_{friction} = 0.2*3*9.8 = 5.88 N $, $ F_{net} = 10 - 5.88 = 4.12 N $, $ a = 4.12 / 3 ≈ 1.37 m/s² $. So $ a $ decreases.
- **Option F (roughen table):** $ \mu $ increases, so $ F_{friction} $ increases. $ F_{net} = F_{pull} - \mu mg $ decreases, so $ a $ decreases.

But the question is which action *reduces* acceleration. Wait, the options are: A, B, C, D, E, F. Wait, the problem says “Click on each action that will reduce the acceleration”. Wait, maybe multiple correct? But the original problem might have a single best answer. Wait, let’s re-read the options:

- A: pull with less force (decreases net force → decreases acceleration)
- C: add an apple (increases mass and friction → decreases acceleration)
- F: roughen table (increases friction → decreases net force → decreases acceleration)

But maybe the intended answer is **A, C, F**? Wait, no—maybe the problem is from a basic physics context where “reducing acceleration” is achieved by decreasing the net force. Let’s check the options again. Wait, the original problem’s diagram shows “FORCE” (pull) and “friction” (opposing).

Wait, the key is: To reduce acceleration, we need to reduce the *net force* (since $ a \propto F_{net} $). Net force is $ F_{pull} - F_{friction} $. So:

- Decreasing $ F_{pull} $ (Option A) reduces net force.
- Increasing $ F_{friction} $ (Option F: roughen table, increases $ \mu $; Option C: add apple, increases $ m $ so $ F_{friction} = \mu mg $ increases) also reduces net force.
- Decreasing mass (Option D) increases acceleration (since $ a = F_{net}/m $, smaller $ m $ → larger $ a $).
- Decreasing friction (Option E) increases net force → increases acceleration.
- Increasing $ F_{pull} $ (Option B) increases net force → increases acceleration.

But if we have to choose the *most probable* action (maybe the simplest), **Option A (pull with less force)** directly reduces the applied force, thus reducing net force and acceleration. Alternatively, **Option C (add an apple)** increases mass (and friction), reducing acceleration. Or **Option F (roughen table)** increases friction, reducing net force.

Wait, maybe the intended answer is **A, C, F** (multiple correct), but let’s check the problem’s wording: “Click on each action that will reduce the acceleration”. So all actions that reduce net force (A, C, F) are correct.

But if we have to pick one, maybe **A** (pull with less force) is the most direct, as it directly reduces the applied force.

# Brief Explanations:
To reduce acceleration, we use $ F_{net} = ma $: acceleration is proportional to net force. Net force is $ F_{pull} - F_{friction} $.

- **Option A:** Pulling with less force decreases $ F_{pull} $, so $ F_{net} $ decreases → acceleration decreases.
- **Option C:** Adding an apple increases mass ($ m $) and friction ($ F_{friction} = \mu mg $ increases), so $ F_{net} $ decreases → acceleration decreases.
- **Option F:** Roughening the table increases friction ($ \mu $ increases), so $ F_{friction} $ increases → $ F_{net} $ decreases → acceleration decreases.

# Answer:
A. pull with less force, C. add an apple to the basket, F. roughen the surface of the table to increase friction

(If only one is required, the most direct is **A. pull with less force**.)

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QUESTION IMAGE

Question

Explanation:

Step 1: Recall Newton’s Second Law

Analyze Each Option:

Answer: