Question

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. (Take g = 10 m/s²)

Hint:

For height traveled in the last second:

• Use \( h_n = u + \frac{1}{2}g(2n - 1) \) for the distance traveled in the \( n \)-th second.
• Multiply the distance by \( m \cdot g \) to calculate the potential energy loss.

Answer 1

Correct Answer: 300

1. Height Traveled in the Last Second: - The distance traveled in the \( n \)-th second of free fall is: \[ h_n = u + \frac{1}{2} g (2n - 1). \] - Here, \( u = 0 \), \( g = 10 \, \text{m/s}^2 \), \( n = 8 \): \[ h_8 = \frac{1}{2} \cdot 10 \cdot (2 \cdot 8 - 1) = \frac{1}{2} \cdot 10 \cdot 15 = 75 \, \text{m}. \]
2. Loss of Potential Energy: - The loss of potential energy is given by: \[ \Delta PE = mgh_n. \] - Substituting \( m = 0.4 \, \text{kg}, \, g = 10 \, \text{m/s}^2, \, \text{and} \, h_8 = 75 \, \text{m} \): \[ \Delta PE = 0.4 \times 10 \times 75 = 300 \, \text{J}. \]

Final Answer: 300 J