Question

A 2 kg object with 100 J potential energy (with respect to the ground) is released from rest from the top of a building, the speed of the object when it hits the ground is 7 m/s 10 m/s 5 m/s 13 m/s

Answer 1

velocity of object just before hitting the ground = 10 m/s

10 m/sExplanation:

here's the solution : -

=》mechanical energy is always constant, when the object is at the top of the building , the object has potential energy but no kinetic energy,

so, at the top of the building the mechanical energy = potential energy = 100 J

but at when it is dropped from the building, the whole potential energy get converted into kinetic energy, just before reaching the bottom .

so, mechanical energy in the object just before hitting the ground = kinetic energy

( potential energy = 0 )

So, kinetic energy = 100 J

now, we know

$kinetic \: energy = \frac{1}{2} \times m {v}^{2}$

=》

$100 = \frac{1}{2} \times 2 \times {v}^{2}$

( since, mass = 2 kg )

=》

$100 = {v}^{2}$

=》

$v = \sqrt{100}$

=》

$v = 10$

so, velocity of object just before hitting the ground = 10 m/s