Question

a block is projected up on smooth fixed inclined plane having angle of incidence 60 degree with speed √6 gh maximum height attained by block is​

Answer 1

Correct Question: A block is projected up on a smooth fixed inclined plane having an angle of inclination 60 degrees with a speed of √6gh. The maximum height attained by the block is​?

The maximum height attained by the block is equal to 3h.

Given:

The angle of inclination of the inclined plane (θ) = 60°

The initial velocity of the block (u) = √(6gh)

To Find:

The maximum height (H) attained by the block.

Solution:

Let the maximum height attained by the block be 'H'.

Let the mass of the block be 'm'.

Assume: Acceleration due to gravity (g) = 10 ms⁻²

→ According to the Law of conservation of energy for a system the total energy of a system remains conserved, it only changes its form from one form to another.

→ In the case of a body which is under the influence of gravity, the mechanical energy which is the sum of potential energy (U) and kinetic energy (K) remains conserved. Hence for a body under the influence of gravity, the total energy which is the sum of potential energy and kinetic energy at any instant always remains constant.

Initial Potential energy of the block (U₁) = 0 (as the block is at ground level)

Initial Kinetic energy of the block (K₁) = 1/2(mu²) = 1/2(m)(√6gh)² = 3mgh

Final Kinetic energy of the block (K₂) = 0 (as the block finally stops)

Final Potential energy of the block (U₂) = mgH

By the law of conservation of energy the total mechanical energy will remain constant:          

                                     ∵ U₁ + K₁ = U₂ + K₂

                                      → 0 + 3mgh = mgH + 0

                                        → 3mgh = mgH

                                              ∴ H = 3h

Hence the maximum height 'H' attained by the block is equal to 3h.

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