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A block is moving on an inclined plane m...

A block is moving on an inclined plane making an angle `45^@` with the horizontal and the coefficient of friction is `mu`. The force required to just push it up the inclined plane is 3 times the force required to just prevent it from sliding down. If we define `N=10mu`, then N is

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AI Generated Solution

To solve the problem, we need to analyze the forces acting on the block on the inclined plane. Let's break it down step by step. ### Step 1: Identify the forces acting on the block 1. **Weight (W)**: The weight of the block acts vertically downward and is given by \( W = mg \), where \( m \) is the mass of the block and \( g \) is the acceleration due to gravity. 2. **Normal Force (N)**: The normal force acts perpendicular to the surface of the incline. 3. **Frictional Force (F_f)**: The frictional force acts parallel to the surface of the incline and opposes the motion. It is given by \( F_f = \mu N \), where \( \mu \) is the coefficient of friction. ### Step 2: Resolve the weight into components ...
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Knowledge Check

  • A brick of mass 2kg just begins to slide down on inclined plane at an angle of 45^(@) with the horizontal. The force of friction will be

    A
    `19.6 sin 45^(@)`
    B
    `19.6cos45^(@)`
    C
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    D
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    A
    `19.6sin45^(@)`
    B
    `19.6cos45^(@)`
    C
    `9.8sin45^(@)`
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  • A block of mass m slides down an inclined plane which makes an angle theta with the horizontal. The coefficient of friction between the block and the plane is mu . The force exerted by the block on the plane is

    A
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    B
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