frq practice #1
name:
scenario #1: a student pulls a rope attached to 30 kg box, causing the box to slide across a rough floor. he uses a motion detector to measure the motion of the box as it is being pulled. assuming he begins pulling the box at time t = 0, and the data for the velocity of the box as a function of time is shown on the graph below.

1. use the velocity vs. time graph to help you answer the questions below.
a. identify the section where the box moves with constant velocity.
b. identify the section where the box is slowing down.
c. identify the section where the box is speeding up.
d. what is the acceleration of the box in section a? (hint: find the slope.)
e. what is the acceleration of the box in section c? (hint: find the slope.)

2. draw a free body diagram for:
| the box during section a: | the box during section b: | the box during section c: |
|---------------------------|---------------------------|---------------------------|
| | | |

3. use the information from the problem and any information collected in questions 1 and 2 to answer the questions below.
a. what is the net force of the box in section a? (hint: use $f_{net} = ma$)
b. what is the net force of the box in section b? (hint: use $f_{net} = ma$)
c. what is the net force of the box in section c? (hint: use $f_{net} = ma$)
d. when is the net force on the box the greatest, section a, b, or c? explain.
e. when is the tension force on the box the greatest, section a, b, or c? explain.

Question

frq practice #1
name:
scenario #1: a student pulls a rope attached to 30 kg box, causing the box to slide across a rough floor. he uses a motion detector to measure the motion of the box as it is being pulled. assuming he begins pulling the box at time t = 0, and the data for the velocity of the box as a function of time is shown on the graph below.

1. use the velocity vs. time graph to help you answer the questions below.
   a. identify the section where the box moves with constant velocity.
   b. identify the section where the box is slowing down.
   c. identify the section where the box is speeding up.
   d. what is the acceleration of the box in section a? (hint: find the slope.)
   e. what is the acceleration of the box in section c? (hint: find the slope.)

2. draw a free body diagram for:
| the box during section a: | the box during section b: | the box during section c: |
|---------------------------|---------------------------|---------------------------|
|                           |                           |                           |

3. use the information from the problem and any information collected in questions 1 and 2 to answer the questions below.
   a. what is the net force of the box in section a? (hint: use $f_{net} = ma$)
   b. what is the net force of the box in section b? (hint: use $f_{net} = ma$)
   c. what is the net force of the box in section c? (hint: use $f_{net} = ma$)
   d. when is the net force on the box the greatest, section a, b, or c? explain.
   e. when is the tension force on the box the greatest, section a, b, or c? explain.

Accepted Answer

### 1a
# Explanation:
## Step1: Recall constant velocity condition
Constant velocity means velocity doesn't change, so the velocity - time graph should be a horizontal line (slope = 0).
## Step2: Identify the section
Looking at the graph, Section B (from $t = 5$ s to $t=7$ s) has a horizontal line, so velocity is constant here.
# Answer:
Section B

### 1b
# Explanation:
## Step1: Recall slowing down condition
Slowing down means velocity is decreasing, so the velocity - time graph has a negative slope (velocity is going from a higher value to a lower value).
## Step2: Identify the section
Section C (from $t = 7$ s to $t = 8$ s) has a negative slope (velocity decreases from 50 m/s to 0 m/s), so the box is slowing down here.
# Answer:
Section C

### 1c
# Explanation:
## Step1: Recall speeding up condition
Speeding up means velocity is increasing, so the velocity - time graph has a positive slope (velocity is going from a lower value to a higher value).
## Step2: Identify the section
Section A (from $t = 0$ s to $t=5$ s) has a positive slope (velocity increases from 0 m/s to 50 m/s), so the box is speeding up here.
# Answer:
Section A

### 1d
# Explanation:
## Step1: Recall acceleration formula
Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v=v_f - v_i$ and $\Delta t=t_f - t_i$.
## Step2: Determine values for Section A
From the graph, at $t_i = 0$ s, $v_i = 0$ m/s; at $t_f=5$ s, $v_f = 50$ m/s.
## Step3: Calculate acceleration
$a=\frac{50 - 0}{5 - 0}=\frac{50}{5}=10$ m/s².
# Answer:
$10$ m/s²

### 1e
# Explanation:
## Step1: Recall acceleration formula
Acceleration $a=\frac{\Delta v}{\Delta t}$, where $\Delta v = v_f - v_i$ and $\Delta t=t_f - t_i$.
## Step2: Determine values for Section C
From the graph, at $t_i = 7$ s, $v_i = 50$ m/s; at $t_f = 8$ s, $v_f=0$ m/s.
## Step3: Calculate acceleration
$a=\frac{0 - 50}{8 - 7}=\frac{- 50}{1}=- 50$ m/s². The negative sign indicates deceleration, but for the magnitude of acceleration (rate of change of velocity), we can also consider the slope. The magnitude of acceleration is $50$ m/s².
# Answer:
$- 50$ m/s² (or a magnitude of $50$ m/s²)

### 2
# Explanation:
## Section A (Speeding up, positive acceleration)
- **Forces on the box**:
    - **Gravitational force ($F_g$)**: Acts downward, $F_g=mg$, where $m = 30$ kg and $g = 9.8$ m/s² (or $10$ m/s² for approximation).
    - **Normal force ($F_N$)**: Acts upward, equal in magnitude to $F_g$ (since there is no vertical acceleration), so $F_N=F_g$.
    - **Tension force ($F_T$)**: Acts to the right (in the direction of motion).
    - **Frictional force ($F_f$)**: Acts to the left (opposite to the direction of motion). Since the box is speeding up, $F_T>F_f$, so the net force $F_{net}=F_T - F_f=ma$ (positive, in the direction of motion).
    - The free - body diagram will have $F_g$ downward, $F_N$ upward, $F_T$ to the right, and $F_f$ to the left, with $F_T$ longer than $F_f$.

## Section B (Constant velocity, acceleration $a = 0$)
- **Forces on the box**:
    - **Gravitational force ($F_g$)**: Acts downward.
    - **Normal force ($F_N$)**: Acts upward, $F_N = F_g$.
    - **Tension force ($F_T$)**: Acts to the right.
    - **Frictional force ($F_f$)**: Acts to the left. Since $a = 0$, $F_{net}=0$, so $F_T=F_f$.
    - The free - body diagram will have $F_g$ downward, $F_N$ upward, $F_T$ to the right, and $F_f$ to the left, with $F_T$ and $F_f$ of equal length.

## Section C (Slowing down, negative acceleration)
- **Forces on the box**:
    - **Gravitational force ($F_g$)**: Acts downward.
    - **Normal force ($F_N$)**: Acts upward, $F_N = F_g$.
    - **Tension force ($F_T$)**: Acts to the right (assuming the student is still pulling, but the box is slowing down).
    - **Frictional force ($F_f$)**: Acts to the left. Since the box is slowing down, $F_f>F_T$, so the net force $F_{net}=F_f - F_T=ma$ (negative, opposite to the direction of motion).
    - The free - body diagram will have $F_g$ downward, $F_N$ upward, $F_T$ to the right, and $F_f$ to the left, with $F_f$ longer than $F_T$.

### 3a
# Explanation:
## Step1: Recall net force formula
The net force formula is $F_{net}=ma$.
## Step2: Determine values for Section A
We know that $m = 30$ kg and from part 1d, $a = 10$ m/s².
## Step3: Calculate net force
$F_{net}=30\ kg\times10\ m/s² = 300$ N.
# Answer:
$300$ N

### 3b
# Explanation:
## Step1: Recall net force formula
The net force formula is $F_{net}=ma$.
## Step2: Determine acceleration for Section B
In Section B, the velocity is constant, so acceleration $a = 0$ m/s².
## Step3: Calculate net force
$F_{net}=30\ kg\times0\ m/s²=0$ N.
# Answer:
$0$ N

### 3c
# Explanation:
## Step1: Recall net force formula
The net force formula is $F_{net}=ma$.
## Step2: Determine acceleration for Section C
From part 1e, $a=- 50$ m/s² (the magnitude of acceleration is $50$ m/s²). We use the magnitude for force calculation (since force is a vector, but the magnitude of net force is $|F_{net}|=m|a|$).
## Step3: Calculate net force
$|F_{net}|=30\ kg\times50\ m/s² = 1500$ N (the negative sign in acceleration just indicates direction, the magnitude of the net force is $1500$ N).
# Answer:
$1500$ N (magnitude)

### 3d
# Explanation:
## Step1: List net forces
- Section A: $F_{netA}=300$ N
- Section B: $F_{netB}=0$ N
- Section C: $|F_{netC}| = 1500$ N
## Step2: Compare the forces
Since $1500\ N>300\ N > 0\ N$, the net force is greatest in Section C.
# Answer:
Section C. Because the magnitude of the net force in Section C ($1500$ N) is greater than in Section A ($300$ N) and Section B ($0$ N).

### 3e
# Explanation:
## Step1: Analyze forces in each section
- **Section A**: $F_{netA}=F_{TA}-F_{fA}=ma_A$, so $F_{TA}=F_{fA}+ma_A$
- **Section B**: $F_{netB}=F_{TB}-F_{fB}=0$ (since $a_B = 0$), so $F_{TB}=F_{fB}$
- **Section C**: $F_{netC}=F_{fC}-F_{TC}=ma_C$ (magnitude), so $F_{TC}=F_{fC}-ma_C$ (assuming $F_{f}$ is constant, or at least $F_{fC}$ is not less than $F_{fA}$ or $F_{fB}$ in a way that would make $F_{TC}$ larger than $F_{TA}$). But in Section A, the box is speeding up, which means that the tension has to overcome friction and also provide the force for acceleration. In Section C, the box is slowing down, so friction is greater than tension. In Section B, tension equals friction. Since $F_{TA}=F_{fA}+ma_A$ and $F_{TB}=F_{fB}$, and if we assume that the frictional force is approximately constant ($F_{fA}\approx F_{fB}\approx F_{fC}$ for a rough floor with the same normal force), then $F_{TA}$ is larger than $F_{TB}$ and $F_{TC}$ because it has to provide the acceleration force in addition to overcoming friction. Wait, no, let's re - consider. Wait, in Section C, the acceleration is negative (deceleration), the net force is $F_{f}-F_{T}=ma$ (magnitude $ma$). In Section A, $F_{T}-F_{f}=ma$. If the frictional force is the same (since normal force is the same, $F_N = mg$, and friction $F_f=\mu F_N$, so $\mu$ and $F_N$ are constant, so $F_f$ is constant). Let $F_f = f$. Then:
    - In Section A: $F_{TA}=f + ma_A$
    - In Section B: $F_{TB}=f$
    - In Section C: $F_{TC}=f - ma_C$ (since $F_f - F_{TC}=ma_C$, so $F_{TC}=f - ma_C$)
Since $ma_A$ and $ma_C$ are positive, $F_{TA}=f + ma_A>f=F_{TB}$ and $F_{TA}=f + ma_A>f - ma_C=F_{TC}$ (because $ma_A$ and $ma_C$ are positive, so $f+ma_A>f - ma_C$). So the tension is greatest in Section A.
Wait, maybe my previous analysis of Section C was wrong. Let's re - do the force analysis. The box is moving to the right in all sections (since velocity is positive). So in Section A: motion to the right, speeding up, so net force to the right ($F_T - F_f=ma$). In Section B: motion to the right, constant velocity, net force zero ($F_T = F_f$). In Section C: motion to the right, slowing down, so net force to the left ($F_f - F_T=ma$). So tension in Section A is $F_T=f + ma$, in Section B is $F_T = f$, in Section C is $F_T=f - ma$. Since $ma$ is positive, $F_{TA}>F_{TB}$ and $F_{TA}>F_{TC}$. So the tension is greatest in Section A.
# Answer:
Section A. Because in Section A, the tension force has to overcome the frictional force and also provide the force required to accelerate the box ($F_T=F_f + ma$), while in Section B, tension only needs to balance friction ($F_T = F_f$) and in Section C, friction is greater than tension ($F_T=F_f - ma$), so the tension in Section A is the greatest.

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QUESTION IMAGE

Question

Explanation:

1a

Step1: Recall constant velocity condition

Step2: Identify the section

Step1: Recall slowing down condition

Step2: Identify the section

Step1: Recall speeding up condition

Step2: Identify the section

Answer:

1b