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3. A 75 Kg block is sliding down a 20 degree incline. The coefficient of friction between the block and the incline is .21. Find the ACCELERATION down the ramp. M = 75 kg = 0.21 20

3. A 75 Kg block is sliding down a 20 degree incline. The coefficient of friction between the block and the incline is .21. Find the ACCELERATION down the ramp. M = 75 kg = 0.21 20

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Solve both questions from both images please. Thank you.

**Problem:** A 75 kg block is sliding down a 20 degree incline. The coefficient of friction between the block and the incline is 0.21. Find the **acceleration** down the ramp. **Diagram Explanation:** - The diagram features a block sliding down an inclined plane. - The mass of the block is labeled as \( M = 75 \, \text{kg} \). - The angle of the incline with the horizontal is \( 20^\circ \). - The coefficient of friction (\( \mu \)) between the block and the plane is given as 0.21. To solve for acceleration, one would typically use the formula: \[ a = g (\sin \theta - \mu \cos \theta) \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), \( \theta \) is the angle of the incline, and \( \mu \) is the coefficient of friction. Replace \( g \), \( \theta \), and \( \mu \) in the formula to find the acceleration.
Transcribed Image Text:**Problem:** A 75 kg block is sliding down a 20 degree incline. The coefficient of friction between the block and the incline is 0.21. Find the **acceleration** down the ramp. **Diagram Explanation:** - The diagram features a block sliding down an inclined plane. - The mass of the block is labeled as \( M = 75 \, \text{kg} \). - The angle of the incline with the horizontal is \( 20^\circ \). - The coefficient of friction (\( \mu \)) between the block and the plane is given as 0.21. To solve for acceleration, one would typically use the formula: \[ a = g (\sin \theta - \mu \cos \theta) \] where \( g \) is the acceleration due to gravity (approximately \( 9.81 \, \text{m/s}^2 \)), \( \theta \) is the angle of the incline, and \( \mu \) is the coefficient of friction. Replace \( g \), \( \theta \), and \( \mu \) in the formula to find the acceleration.
**Problem Statement:** Rachel pushes down on a stationary 33 kg block with a force of 330 N at an angle of 28° below the horizontal. If the coefficient of friction between the block and the ground is 0.17, find the acceleration of the block. **Diagram Explanation:** The diagram shows a block on a horizontal surface. A force vector is drawn from the block at a 28° angle below a horizontal line, indicating the direction of the force applied by Rachel. **Solution Steps:** To solve this problem, follow these steps: 1. **Resolve the Force into Components:** - Calculate the horizontal component of the force (\( F_{\text{x}} \)) using \( F_{\text{x}} = F \cdot \cos(\theta) \). - Calculate the vertical component of the force (\( F_{\text{y}} \)) using \( F_{\text{y}} = F \cdot \sin(\theta) \). 2. **Determine the Normal Force:** - Calculate the normal force (\( N \)) on the block, which is affected by the weight of the block and the vertical component of the applied force: \[ N = mg + F_{\text{y}} \] where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²), and \( F_{\text{y}} \) is the downward component of the applied force. 3. **Calculate the Frictional Force:** - The frictional force (\( f_{\text{friction}} \)) is given by: \[ f_{\text{friction}} = \mu \cdot N \] where \( \mu \) is the coefficient of friction. 4. **Net Force and Acceleration:** - Determine the net horizontal force (\( F_{\text{net}} \)) acting on the block by subtracting the frictional force from the horizontal component of the applied force: \[ F_{\text{net}} = F_{\text{x}} - f_{\text{friction}} \] - Calculate the acceleration (\( a \)) using Newton's second law: \[ a = \frac{F_{\text{net}}}{m} \] Solve each step to find the
Transcribed Image Text:**Problem Statement:** Rachel pushes down on a stationary 33 kg block with a force of 330 N at an angle of 28° below the horizontal. If the coefficient of friction between the block and the ground is 0.17, find the acceleration of the block. **Diagram Explanation:** The diagram shows a block on a horizontal surface. A force vector is drawn from the block at a 28° angle below a horizontal line, indicating the direction of the force applied by Rachel. **Solution Steps:** To solve this problem, follow these steps: 1. **Resolve the Force into Components:** - Calculate the horizontal component of the force (\( F_{\text{x}} \)) using \( F_{\text{x}} = F \cdot \cos(\theta) \). - Calculate the vertical component of the force (\( F_{\text{y}} \)) using \( F_{\text{y}} = F \cdot \sin(\theta) \). 2. **Determine the Normal Force:** - Calculate the normal force (\( N \)) on the block, which is affected by the weight of the block and the vertical component of the applied force: \[ N = mg + F_{\text{y}} \] where \( m \) is the mass of the block, \( g \) is the acceleration due to gravity (approximated as 9.8 m/s²), and \( F_{\text{y}} \) is the downward component of the applied force. 3. **Calculate the Frictional Force:** - The frictional force (\( f_{\text{friction}} \)) is given by: \[ f_{\text{friction}} = \mu \cdot N \] where \( \mu \) is the coefficient of friction. 4. **Net Force and Acceleration:** - Determine the net horizontal force (\( F_{\text{net}} \)) acting on the block by subtracting the frictional force from the horizontal component of the applied force: \[ F_{\text{net}} = F_{\text{x}} - f_{\text{friction}} \] - Calculate the acceleration (\( a \)) using Newton's second law: \[ a = \frac{F_{\text{net}}}{m} \] Solve each step to find the
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