Question

A particle is released from height \(s\) above the surface of the earth. At a certain height its K.E is 3 times of P.E. The height from the surface of the earth and the speed of the particle at the instant are respectively:

• A. \( \frac{s}{4}, \, \sqrt{\frac{3gs}{2}} \)
• B. \( \frac{s}{2}, \, \sqrt{\frac{3gs}{2}} \)
• C. \( \frac{s}{2}, \, \sqrt{\frac{3gs}{2}} \)
• D. \( \frac{s}{4}, \, \sqrt{\frac{3gs}{2}} \)

Hint:

In problems involving energy conservation, use the relationship between K.E and P.E to find the velocity and height at certain points.

Answer 1

The Correct Option is A

Let the total mechanical energy be the sum of K.E and P.E: \[ E_{\text{total}} = \text{K.E} + \text{P.E} \] The total mechanical energy is conserved, so it is constant throughout the motion of the particle. We are given that the K.E is 3 times the P.E at a certain point: \[ \text{K.E} = 3 \cdot \text{P.E} \] At that point, the total mechanical energy is: \[ E_{\text{total}} = 3 \cdot \text{P.E} + \text{P.E} = 4 \cdot \text{P.E} \] Thus, at that height, the potential energy is \( \frac{1}{4} \) of the total mechanical energy, and the speed of the particle can be obtained from energy conservation principles. Using the energy conservation and solving, we get the height from the surface and the speed as \( \frac{s}{4} \) and \( \sqrt{\frac{3gs}{2}} \), respectively.