Answer
$K.E=1.2J$
Work Step by Step
The kinetic energy is given as:
$K.E=\frac{1}{2}mv^2$
Also,
$v=x_m\omega$
Combining these two equations,
$K.E=\frac{1}{2}m{x_m}^2\omega^2$..........eq(1)
We also know that,
$\omega=\frac{2\pi}{T}=\frac{2(3.1416)}{0.04}=157\frac{rad}{s}$
We plug in the known values in eq(1) to obtain;
$K.E=\frac{1}{2}(0.020)(0.07)^2(157)^2=1.2J$
