Find the force on an object on an inclined plane

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SUMMARY

The discussion centers on calculating the force acting on an object projected up an inclined plane at an angle θ, specifically when the object reaches the top of a sphere with radius R. The minimum velocity required to reach the top is derived as √(2g[R(1-cos θ) + L sin θ). The user initially miscalculates the force using the formula (mV²)/R, leading to confusion regarding the object's velocity at the top. The correct approach involves applying conservation of energy, confirming that the object does not arrive with half the initial velocity but retains sufficient kinetic energy to maintain contact with the incline.

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  • Understanding of Newton's laws of motion
  • Familiarity with circular motion dynamics
  • Knowledge of conservation of energy principles
  • Basic trigonometry related to inclined planes
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  • Study the principles of circular motion and centripetal force
  • Learn about energy conservation in mechanical systems
  • Explore the dynamics of objects on inclined planes
  • Investigate the effects of varying angles on motion and forces
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Physics students, educators, and anyone interested in mechanics, particularly those studying forces on inclined planes and circular motion dynamics.

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Homework Statement



A body is projected up an inclined plane with a speed Vo. The inclination of the incline is [tex]\theta[/tex]. The surface of th incline is a tangent to a circle of radius R. Find the force acting on th object when it is at the top of the sphere provied that Vo= 2*the minimum velocity need to reach the top of the sphere.
The length of incline = L
Assume that the block does not lose contact with the track.

The Attempt at a Solution


The minimum velocity required, as I calculated was:
[tex]\sqrt{2g[R(1-cos \theta)+ L sin \theta}[/tex] which was correct.
But the Force I calculated was diffrent.

At the top of the phere, the velocity will become Vo/2. Why oesnt (mV^2)/R give me the correct answer? as the body perform circular motion. Help me please.
 
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The minimum velocity required to reach the top satisfies (1/2)mv0^2=mgh (where h is the total height of the top of the sphere) - use conservation of energy. If you leave with twice that velocity your energy at the top is (1/2)m(2*v0)^2-mgh. Equate that to kinetic energy at the top and solve for velocity. You DON'T arrive with half the initial velocity.
 

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