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A body of mass 20 kg is dropped from a height of 100 m. Find its kinetic energy and potential energy after 1 sec, 2 sec, 3 sec.

(a) KE=0J,PE=2000J at 1 sec
(b) KE=200J,PE=1800J at 1 sec
(c) KE=400J,PE=1600J at 1 sec
(d) KE=2000J,PE=0J at 1 sec

1 Answer

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Final answer:

Using the equations for gravitational potential energy and kinetic energy, we find that after 1 second, none of the given options accurately reflect the kinetic energy (960 J) or potential energy (18619.8 J) of the falling body.

Step-by-step explanation:

The question is asking to calculate the kinetic energy (KE) and potential energy (PE) of a 20 kg mass dropped from 100 m after 1, 2, and 3 seconds. To find these energies, we can use the equations for gravitational potential energy (PE = mgh) and kinetic energy (KE = 0.5mv²), knowing that g (acceleration due to gravity) is approximately 9.8 m/s².

After 1 second:


  • The mass would have fallen with a velocity v = g * t = 9.8 m/s * 1 s = 9.8 m/s.

  • Its subsequent kinetic energy would be KE = 0.5 * m * v² = 0.5 * 20 kg * (9.8 m/s)² = 960 J.

  • The height h from which it fell can be calculated using h = 0.5 * g * t² = 0.5 * 9.8 m/s² * (1 s)² = 4.9 m.

  • Therefore, remaining height is 100 m - 4.9 m = 95.1 m and PE = m * g * height = 20 kg * 9.8 m/s² * 95.1 m = 18619.8 J.

After 2 and 3 seconds, the process for calculating KE and PE is the same, though with different velocities and heights corresponding to each time interval.

None of the options (a), (b), (c), or (d) are correct for the kinetic energy and potential energy after 1 second, as the correct values are KE = 960 J and PE = 18619.8 J.

User Nikita Leonov
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