Answer
Please see the work below.
Work Step by Step
(a) We know that
$U=\frac{1}{2}Kx^2$
$\implies U=\frac{1}{2}K(0)^2=0J$
$K.E=\frac{1}{2}mv_i^2$
We plug in the known values to obtain:
$K.E=\frac{1}{2}(1.4Kg)(0.95m/s)^2=0.63175J$
Total energy is given as
$E=K.E+U=0.63175+0=0.63175J$
(b) We know that
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtian:
$\implies U=\frac{1}{2}(734N/m)(1\times 10^{-2}m)^2=0.0367J$
Now the kinetic energy is given as
$K.E=E-U$
We plug in the known values to obtain:
$K.E=0.63175J-0.0367J=0.59505J$
(c) We know that
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtian:
$\implies U=\frac{1}{2}(734N/m)(0.02m)^2=0.1468J$
Now the kinetic energy is given as
$K.E=E-U$
We plug in the known values to obtain:
$K.E=0.63175J-0.1468J=0.48495J$
(d) We know that
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtian:
$\implies U=\frac{1}{2}(734N/m)(0.03m)^2=0.3303J$
Now the kinetic energy is given as
$K.E=E-U$
We plug in the known values to obtain:
$K.E=0.63175J-0.3303J=0.30145J$
(e) We know that
$U=\frac{1}{2}Kx^2$
We plug in the known values to obtian:
$\implies U=\frac{1}{2}(734N/m)(0.04m)^2=0.5872J$
Now the kinetic energy is given as
$K.E=E-U$
We plug in the known values to obtain:
$K.E=0.63175J-0.5872J=0.04455J$
