A 0.4 kg mass takes 8 s to reach ground when dropped from a certain height ' P ' above surface of earth. The loss of potential energy in the last second of fall is J.

JEE Mains - Potential Energy Loss in Free Fall

Question: A mass \( m = 0.4\,\text{kg} \) takes \( 8\,\text{s} \) to reach the ground when dropped from a certain height \( P \) above the surface of the Earth. Given \( g = 10\,\text{m/s}^2 \), what is the loss of potential energy in the last second of the fall (in Joules)?

[29-Jan-2023 Shift 1]

Select the correct answer:

(A) \(300\,\text{J}\)

(B) \(320\,\text{J}\)

(C) \(400\,\text{J}\)

(D) \(500\,\text{J}\)

Solution

For an object in free fall, the distance fallen in time \( t \) is given by: \[ s = \frac{1}{2}gt^2. \]

The total distance fallen in \( 8\,\text{s} \) is: \[ s_8 = \frac{1}{2} \times 10 \times 8^2 = 5 \times 64 = 320\,\text{m}. \] Similarly, in \( 7\,\text{s} \): \[ s_7 = \frac{1}{2} \times 10 \times 7^2 = 5 \times 49 = 245\,\text{m}. \]

Therefore, the distance covered in the last second (from \( t=7\,\text{s} \) to \( t=8\,\text{s} \)) is: \[ \Delta s = s_8 - s_7 = 320 - 245 = 75\,\text{m}. \]

The loss of potential energy in that last second is: \[ \Delta PE = m\,g\,\Delta s = 0.4 \times 10 \times 75 = 300\,\text{J}. \]