Question

A body of mass 1kg is moving in a vertical circular path of radius 1m .The diff. Between the k.E at its highest and lowest point is

Answer 1

Answer:

20

Explanation:

at highest point its potential energy will be m*g*2r

at lowest point its potential energy will be m*g*0

so,dif. in k.e. will be

mg2r-mg0

=1*10*2*1

=20

so, final ans is 20

Answer 2

The difference between the k.E at its highest and lowest point is 19 .6 J.

Explanation:

Given mass of the body, m = 1 kg.

Radius of the circle, r = 1 m.

So the highest point circular path, h = 2 m.

At highest point, the kinetic energy converted to potential energy, therefore

K.E = P.E = mgh

$K.E = 1\times9.8\times2m$

K.E = 19. 6 J.

At lowest point the kinetic energy

k.e = 0 because h = 0.

Thus, the difference between the k.E at its highest and lowest point is 19 .6 J.

#Learn More:

Topic: Kinetic energy and potential energy

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