Coefficient of kinetic friction between the mass and the incline 0.2. Find the minimum work done to slide the mass from the bottom to the top of the incline. F, = 50 N 10 kg h = 4 m 0 = 30

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**Problem Statement:**

Coefficient of kinetic friction between the mass and the incline is 0.2. Find the minimum work done to slide the mass from the bottom to the top of the incline.

**Diagram Explanation:**

- The diagram shows a 10 kg block on an inclined plane with an angle of \( \theta = 30^\circ \).
- The height of the incline is given as \( h = 4 \, \text{m} \).
- A force, \( f_1 = 50 \, \text{N} \), is applied parallel to the incline.
- The coefficient of kinetic friction is 0.2.

**Physics Concepts:**

The work done to move an object up an incline is calculated considering gravitational forces, frictional forces, and any applied force. The work done \( W \) can be expressed as:

\[ W = mgh + f_f d \]

where:
- \( m \) is mass of the object,
- \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \),
- \( h \) is the height of the incline,
- \( f_f \) is the frictional force,
- \( d \) is the distance along the incline.

The distance \( d \) can be found using trigonometry:
- \( d = \frac{h}{\sin(\theta)} \).

The frictional force \( f_f \) is calculated as:
\[ f_f = \mu_k \cdot N \]
Where \( N \) is the normal force.

The normal force \( N \) on an incline is:
\[ N = mg \cos(\theta) \]

Therefore, the work done against friction is:
\[ W_{\text{friction}} = f_f \cdot d = \mu_k \cdot mg \cos(\theta) \cdot d \]

Finally, add these components to find the total work done to overcome gravity and friction.
Transcribed Image Text:**Problem Statement:** Coefficient of kinetic friction between the mass and the incline is 0.2. Find the minimum work done to slide the mass from the bottom to the top of the incline. **Diagram Explanation:** - The diagram shows a 10 kg block on an inclined plane with an angle of \( \theta = 30^\circ \). - The height of the incline is given as \( h = 4 \, \text{m} \). - A force, \( f_1 = 50 \, \text{N} \), is applied parallel to the incline. - The coefficient of kinetic friction is 0.2. **Physics Concepts:** The work done to move an object up an incline is calculated considering gravitational forces, frictional forces, and any applied force. The work done \( W \) can be expressed as: \[ W = mgh + f_f d \] where: - \( m \) is mass of the object, - \( g \) is the acceleration due to gravity \( (9.8 \, \text{m/s}^2) \), - \( h \) is the height of the incline, - \( f_f \) is the frictional force, - \( d \) is the distance along the incline. The distance \( d \) can be found using trigonometry: - \( d = \frac{h}{\sin(\theta)} \). The frictional force \( f_f \) is calculated as: \[ f_f = \mu_k \cdot N \] Where \( N \) is the normal force. The normal force \( N \) on an incline is: \[ N = mg \cos(\theta) \] Therefore, the work done against friction is: \[ W_{\text{friction}} = f_f \cdot d = \mu_k \cdot mg \cos(\theta) \cdot d \] Finally, add these components to find the total work done to overcome gravity and friction.
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