Question

An object of mass 2kg is released from height 100 m what is the change in P.E.after 2 sec & K.E.of object t = 2sec?​

Answer 1

Answer:

For a freely falling body,

v=gt

where g is acceleration due to gravity and t is time. t=2s

g=10 m/s²

v=gt

So v=10*2

= 20 m/s

(Thus in 2s body would travel a distance of 40 m from release point)

Kinetic Energy K=1/2 mv²

m=2kg v=20m/s

So K= 1/2* 2 *20² = 1* 400 = 400 J

Thus the Kinetic Energy of the body would be 400 J at the end of 2s.

Explanation:

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Answer 2

Explanation:

In 2s,v=20−10×2=0

As, K.E= 2/1mv^2

So, kinetic energy at the highest point will be zero.