250 g Ftens = ↓ ↑ μ = 0.10 ΣFsystem = a system Ftension (force of tension) = asystem (acceleration of the system) = ΣΕ ZF system (net force of the system) = N 50 g m/s/s N

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g= m/s^2
### Physics Problem: Pulley System with Friction

#### Diagram Description:
The diagram shows a pulley system with two masses and a coefficient of friction. 

- There is a block with a mass of 250 grams on a flat surface with a coefficient of friction (μ) of 0.10.
- This block is connected by a string to another block with a mass of 50 grams hanging freely.
- The string passes over a pulley, and the tension in the string is represented by \( F_{\text{tens}} \).

#### Variables to Determine:
1. **Force of Tension ( \( F_{\text{tens}} \) )**:
   \[
   F_{\text{tens}} = \_\_\_\_\_\_\_\_ \text{N}
   \]

2. **Net Force of the System ( \( \sum F_{\text{system}} \) )**:
   \[
   \sum F_{\text{system}} = \_\_\_\_\_\_\_\_
   \]

3. **Acceleration of the System ( \( a_{\text{system}} \) )**:
   \[
   a_{\text{system}} = \_\_\_\_\_\_\_\_ \text{m/s}^2
   \]

#### Additional Information:
- **Mass of block on surface**: 250 g
- **Mass of hanging block**: 50 g
- **Coefficient of friction (μ)**: 0.10

### Equations to Use:

1. **Newton's Second Law for the System**:
   \[
   \sum F = ma
   \]
2. **Force of Friction**:
   \[
   F_{\text{friction}} = \mu \cdot F_{\text{normal}}
   \]

### Steps to determine the variables:

1. **Calculate the Force of Friction**:
   - Normal force for the 250g block is equal to its weight:
     \[
     F_{\text{normal}} = m \cdot g = 0.250 \text{ kg} \cdot 9.8 \text{ m/s}^2
     \]
   - Force of Friction:
     \[
     F_{\text{friction}} = \mu \cdot F_{\text{normal}}
     \]

2. **Determine the Net Force**:
   - The net force exert
Transcribed Image Text:### Physics Problem: Pulley System with Friction #### Diagram Description: The diagram shows a pulley system with two masses and a coefficient of friction. - There is a block with a mass of 250 grams on a flat surface with a coefficient of friction (μ) of 0.10. - This block is connected by a string to another block with a mass of 50 grams hanging freely. - The string passes over a pulley, and the tension in the string is represented by \( F_{\text{tens}} \). #### Variables to Determine: 1. **Force of Tension ( \( F_{\text{tens}} \) )**: \[ F_{\text{tens}} = \_\_\_\_\_\_\_\_ \text{N} \] 2. **Net Force of the System ( \( \sum F_{\text{system}} \) )**: \[ \sum F_{\text{system}} = \_\_\_\_\_\_\_\_ \] 3. **Acceleration of the System ( \( a_{\text{system}} \) )**: \[ a_{\text{system}} = \_\_\_\_\_\_\_\_ \text{m/s}^2 \] #### Additional Information: - **Mass of block on surface**: 250 g - **Mass of hanging block**: 50 g - **Coefficient of friction (μ)**: 0.10 ### Equations to Use: 1. **Newton's Second Law for the System**: \[ \sum F = ma \] 2. **Force of Friction**: \[ F_{\text{friction}} = \mu \cdot F_{\text{normal}} \] ### Steps to determine the variables: 1. **Calculate the Force of Friction**: - Normal force for the 250g block is equal to its weight: \[ F_{\text{normal}} = m \cdot g = 0.250 \text{ kg} \cdot 9.8 \text{ m/s}^2 \] - Force of Friction: \[ F_{\text{friction}} = \mu \cdot F_{\text{normal}} \] 2. **Determine the Net Force**: - The net force exert
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