Question

A body of mass 2 kg is dropped from a height of 1 m. Its
kinetic energy as it touches the ground is
​

Answer 1

Answer:

19.6J

Explanation:

Initially only Potential energy was present but when it reaches the ground only kinetic energy remains.

Since the total energy is conserved,

PE = KE

mgh = 1/2 mv2

2×9.8×1= KE

KE = 19.6J

Hope this helps

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Answer 2

Answer:

the problem can be done in number of ways.. [ assume g = 10 , for simplicity]

first method = use v^2 = u^2 + 2gh

v^ 2 = 0 + 2 × 10 × 1

v^ 2 = 20

therefore K.E = 1/2 × m× v^2

= 1/2 × 2 × 20 = 20

second method = the potential energy at the point where mass is being released is = mgh

i.e 2× 10 × 1. = 20 , now this potential energy will get convert to kinetic energy.. so kinetic energy will be also 20..

hope this helps