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)
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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.
To solve this problem, we need to calculate the loss of potential energy in the last second of the fall for a 0.4 kg mass dropped from a height. First, find the height 'h' from which the object falls. This mass takes 8 seconds to reach the ground. It implies the total time 't' is 8 seconds. Using the equation of motion for free fall: h = 0.5 × g × t2 Substitute g = 10 m/s2 and t = 8 s: h = 0.5 × 10 × 82 = 0.5 × 10 × 64 = 320 m The total potential energy at the start is given by: PEtotal = m × g × h = 0.4 × 10 × 320 = 1280 J Now, determine the height covered in the last second. Calculate the velocity just before the last second using: v7 = g × t = 10 × 7 = 70 m/s Use the velocity at 7 seconds to find the distance fallen in the 8th second: Distance = v7 × t + 0.5 × g × t2 = 70 × 1 + 0.5 × 10 × 1 × 1 = 70 + 5 = 75 m Thus, the height at the beginning of the last second is 75 m below the initial 320 m, which is: 320 m - 75 m = 245 m Calculate the potential energy at this point: PE7 = m × g × height = 0.4 × 10 × 245 = 980 J Loss of potential energy in the last second is: 1280 J - 980 J = 300 J Thus, the loss of potential energy in the last second of fall is 300 J, which fits the expected range of (300,300).
