A body of mass 1 kg lies on smooth inclined plane the block of mass m ...
Analysis of Net Normal Reaction on Block
Given Data:
- Mass of the body (m) = 1 kg
- Applied force (F) = 10 N
- Inclined plane is smooth
Free Body Diagram:
- The forces acting on the block are gravity (mg) acting downward, normal reaction (N) perpendicular to the inclined plane, and the applied force (F) acting horizontally.
- The component of the weight perpendicular to the plane is mgcosθ, where θ is the angle of inclination.
Equations of Motion:
- The net force acting on the block along the incline is F - mg sinθ = ma, where a is the acceleration of the block.
- The normal force N is acting perpendicular to the incline and can be calculated using N - mgcosθ = 0, as the block is not moving perpendicular to the plane.
Calculating Normal Reaction:
- From the equation N - mgcosθ = 0, we get N = mgcosθ.
- Substituting the values, N = (1 kg)(9.8 m/s^2)cosθ.
- Depending on the inclination angle θ, you can calculate the normal reaction N.
Conclusion:
- The magnitude of the net normal reaction on the block depends on the angle of inclination of the plane.
- By using the equations of motion and resolving forces, you can determine the normal reaction acting on the block in this scenario.
A body of mass 1 kg lies on smooth inclined plane the block of mass m ...
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