Box A weighing 12 pounds is being dragged at the speed of 3 ft/s. Box B weighs 25 pounds. The KC for box A is 0.3. Please Find The tension of the cable The velocity of box A t=2 s B

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### Problem Statement:

Box A weighing 12 pounds is being dragged at the speed of 3 ft/s. Box B weighs 25 pounds. The kinetic coefficient (KC) for box A is 0.3. Please find:

- The tension of the cable
- The velocity of box A at \( t = 2 \) s


### Diagram Explanation:

- The diagram illustrates a physical setup where two boxes, A and B, are interconnected via a pulley system.
- **Box A** is positioned on a horizontal surface and connected to a cable.
- **Box B** is suspended vertically below the pulley.
- The system is designed such that the movement of Box A horizontally relates to the vertical movement of Box B through the pulley.

### Instructions:

To solve for the tension in the cable and the velocity of Box A at time \( t = 2 \) seconds, apply the following physics principles:

#### 1. Newton's Second Law:

For Box A (on the horizontal plane, taking friction into account):
\[ F_{\text{net\_A}} = m_A \cdot a_A \]
\[ T - f_{\text{friction}} = m_A \cdot a_A \]
where \( f_{\text{friction}} = \mu_k \cdot N \)
\(\mu_k\) is the kinetic friction coefficient (0.3), and \( N \) is the normal force (equal to the weight of Box A).

For Box B (moving vertically):
\[ F_{\text{net\_B}} = m_B \cdot a_B \]
\[ m_B \cdot g - T = m_B \cdot a_B \]

#### 2. Friction Force Calculation:

\[ f_{\text{friction}} =  \mu_k \cdot m_A \cdot g \]
where \( g \) is the acceleration due to gravity (typically \( 32.2 \) ft/s² in imperial units).

#### 3. Equating Accelerations:

Since the system is connected via a pulley, the accelerations \( a_A \) and \( a_B \) will be the same:
\[ a_A = a_B \]

#### 4. Tension Calculation:

Solve the above set of equations to find the tension \( T \) in the cable.

#### 5. Velocity Calculation:

Determine the velocity of Box A at \( t = 2 \) seconds using
Transcribed Image Text:### Problem Statement: Box A weighing 12 pounds is being dragged at the speed of 3 ft/s. Box B weighs 25 pounds. The kinetic coefficient (KC) for box A is 0.3. Please find: - The tension of the cable - The velocity of box A at \( t = 2 \) s ### Diagram Explanation: - The diagram illustrates a physical setup where two boxes, A and B, are interconnected via a pulley system. - **Box A** is positioned on a horizontal surface and connected to a cable. - **Box B** is suspended vertically below the pulley. - The system is designed such that the movement of Box A horizontally relates to the vertical movement of Box B through the pulley. ### Instructions: To solve for the tension in the cable and the velocity of Box A at time \( t = 2 \) seconds, apply the following physics principles: #### 1. Newton's Second Law: For Box A (on the horizontal plane, taking friction into account): \[ F_{\text{net\_A}} = m_A \cdot a_A \] \[ T - f_{\text{friction}} = m_A \cdot a_A \] where \( f_{\text{friction}} = \mu_k \cdot N \) \(\mu_k\) is the kinetic friction coefficient (0.3), and \( N \) is the normal force (equal to the weight of Box A). For Box B (moving vertically): \[ F_{\text{net\_B}} = m_B \cdot a_B \] \[ m_B \cdot g - T = m_B \cdot a_B \] #### 2. Friction Force Calculation: \[ f_{\text{friction}} = \mu_k \cdot m_A \cdot g \] where \( g \) is the acceleration due to gravity (typically \( 32.2 \) ft/s² in imperial units). #### 3. Equating Accelerations: Since the system is connected via a pulley, the accelerations \( a_A \) and \( a_B \) will be the same: \[ a_A = a_B \] #### 4. Tension Calculation: Solve the above set of equations to find the tension \( T \) in the cable. #### 5. Velocity Calculation: Determine the velocity of Box A at \( t = 2 \) seconds using
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