A block of mass M slides down a frictionless plane inclined at an angle (theta) with the horizontal. The normal reaction force exerted by the plane on the block is

Respuesta :

Answer:

Mg Cos theta

Explanation:

Choosing a coordinate system parallel and perpendicular to the plane, since the acceleration perpendicular to the plane is 0, the net force perpendicular to the plane is also 0. Therefore, the magnitude of the normal force equals the magnitude of the y component of the weight.

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The normal reaction force exerted by the plane on the block is [tex]mgcos\theta[/tex]

For a block of mass M on an inclined plane, the normal reaction acting on the plane is always equal to the weight of that object

The formula for calculating the weight of the object is expressed according to the formula:

W = mg

m is the mass of the object

g is the acceleration due to gravity

Since the body is inclined at an angle theta, the weight will be resolved along the horizontal to have [tex]W = mgcos\theta[/tex]

Since the normal reaction force is equal to the weight, hence [tex]W=R=mgcos\theta[/tex]

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