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A block initially at rest slides down a frictionless inclined plane from a height h. It reaches the bottom of the inclined plan with a final speed v. To attain a final speed 4v, it should fall from a height

A.5h
B.10h
C.16h
D.26h

1 Answer

5 votes

Final answer:

To achieve a final speed of 4v, the block must slide down from a height of 16h, as the potential energy required is 16 times greater to attain the kinetic energy corresponding to the speed 4v.

Step-by-step explanation:

When a block slides down a frictionless inclined plane, its potential energy is converted into kinetic energy. According to the energy conservation principle, the potential energy at height h is equal to the kinetic energy at the bottom of the incline.

The potential energy (PE) can be expressed as PE = mgh, where m is the mass of the block, g is the acceleration due to gravity, and h is the height. The kinetic energy (KE) at the bottom is KE = 0.5 * m * v^2, where v is the final speed of the block.

For the final speed to be 4v, the final kinetic energy would need to be 16 times greater because kinetic energy is proportional to the square of the speed (16 * 0.5 * m * v^2). Therefore, the height from which the block must fall to attain a final speed of 4v is 16h, matching the energy proportionality.

The correct answer is C. 16h.

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