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

Name: g= m/s^2 ### Physics Problem: Pulley System with Friction#### Diagram
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Description: 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 fri

Given Two blocks are connected by a string as shown in the figure We have to calculate the Net force…

Answered: 250 g Ftens = ↓ ↑ μ = 0.10 ΣFsystem = a…

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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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Newton's First Law of Motion

If a very smooth ball is rolled down an inclined surface with zero friction, the ball will continue to roll indefinitely for an infinite time, in an attempt to reach to the same height on the opposite hill!

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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

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