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Motion in smooth curved surface in vertical plane.

A block of mass m: I.62 kg slides down a frictionless incline (attached). The block is released a height h : 3.91m above the bottom of the loop.

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2-25.
A block of mass m: I.62 kg slides down a frictionless incline (Figure 2-A). The block is released a height h : 3.91m above the bottom of the loop.

(a) What is the force of the inclined track on the block at the bottom (point A)?

The track in frictionless and hence the law of conservation of mechanical energy is applicable to the block. According to this law here if the velocity of the block at A is vA then
Gain in kinetic energy = loss in potential energy

(see attached file for equations)

Now as the block is moving on a circular track it is having a centripetal acceleration and requires a centripetal force to provide it. If the normal reaction of the track at point A then the resultant force of the weight of the block mg and the normal reaction N at A will provide necessary centripetal force and hence we get

(see attached file for equations)

Or the force exerted by the track on the block at point A will be

(see attached file ...

Solution Summary

The solution calculate velocity, potential and kinetic energies, and force of reaction at a point of a block on the curved smooth path in vertical plane.

$2.19