Question

The circular blade of a power saw has kinetic energy 55 J .

Part A

If its rotation rate drops to half, what's its new kinetic energy?

Express your answer to two significant figures and include the appropriate units.

Answer 1

Here ,

initial kinetic energy , Ki = 55 J

let the final kinetic energy = Kf

NOw, let the initial rate of rotation is w1

and final rate of rotation is w2

as the rate of rotation is halved

w2 = w1/2

Initital kinetic energy , Ki = 0.5 * I * w1^2

final kinetic energy , Kf = 0.5 * I * w2^2

Kf = 0.5 * I * (w1/2)^2

Kf = 0.25 * 0.5 * I * w1^2

Kf = 0.25 * 55

Kf = 13.75 J

the new kinetic energy of the blaade is 13.75 J