Question

A block is placed on a rough incline plane having an inclination angle theta. Now, the angle is increased from 0 to 90, then the contact force between the block and the plane. Remains constant first increases then decreases First decreases then increases first remains constant then decreases

Answer 1

Given:

A block is placed on a rough incline plane having an inclination angle theta. Now, the angle is increased from 0 to 90°.

To find:

Variation in Contact force.

Calculation:

When an object is placed on an inclined plane (having an inclination angle $\theta$ ) , the weight of the object can be distributed into two perpendicular components.

The cosine component which is present perpendicular to the inclined plane constitutes the the variation in the contact force.

Let contact force be F

$F = mg \cos( \theta)$

For constant m and g ;

$= > \: F \propto \: \cos( \theta)$

Since value of $\cos(\theta)$ decreases in the range of 0° to 90° , similar variation will be observed in contact force.

Hence , contact force will decrease if we increase the angle from 0° to 90°.

Answer 2

Answer:

The contact force here is friction.

The contact force here is friction.Intitially static friction would be there which would be equal to mgsinθ ,as angle of inclination increases sinθ will increase.It will increase till it reaches its maximum value i.e umgcosθ (max static friction) .after that it will remain same i.e umgcosθ but

tic friction) .after that it will remain same i.e umgcosθ but θ will continue to increase as a result cosθ would decrease decreasing the friction applied .

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