Question

a body of mass 2 kg Falls freely under g
ravity from a height of 200 m what will be its kinetic energy during the fall at the ends of 5 second​

Answer 1

Answer:

400J

Explanation:

when the body is released from a height of 2 meters then all the potential energy will be converted to kinetic energy.

assuming base as a reference

mgh+0=K.E.+0

so K.E.=400J

Answer 2

Kinetic energy after 5 sec is 4225 J

Explanation:

Given,

Mass = 2 kg

Acceleration = acceleration due to gravity (g)

                      = 10 ms⁻² (approximately)

Time = 5 sec

Height = 200 m

Kinetic energy during fall is $\frac{1}{2} mv^{2}$

By using equations of motion we can calculate the final velocity of body

=> $s=ut+\frac{1}{2} at^{2}$

=> $200= u(5)+\frac{1}{2}10(5^{2} )$                 .........................(1)

Also by using v = u + at

=> v = u + 10 (5)                            ..............................(2)

By solving equation (1) and (2) we get

=> v= 65 m/s

K.E. = $\frac{1}{2} 2 * 65^{2}$

=> K.E. = 4225 J

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