91Ó°ÊÓ

The given data can be listed below as,

- The mass of baseball is, $m=0.145\mathrm{kg}$

- The speed of baseball is, $v=17\mathrm{m}/\mathrm{s}$

Energy is equivalent to the mass of a particle at rest in an inertial frame of reference, energy equal to its rest mass multiplied by the square of the speed of light.

The relation of rest energy is expressed as,

$E_0=m{c}^2$

(i)

Here $E_0$is the rest energy, $m$is the mass of the baseball and $c$is the speed of light. Substitute $0.145\mathrm{kg}$for $m$and $3×{10}^8\mathrm{m}/\mathrm{s}$for $c$in the equation (i).

$E_0=0.145\mathrm{kg}×{\left(3×{10}^8\mathrm{m}/\mathrm{s}\right)}^2$

Hence the rest energy is, $1.305×{10}^{\text{'}8}\mathrm{J}$.

The relation of kinetic energy is expressed as,

$K-\frac{1}{2}m{v}^2$

Here $K$is the kinetic energy, $m$is the mass of the baseball and $v$is the speed of baseball.

Substitute $0.145\mathrm{kg}$and $17\mathrm{m}/\mathrm{s}$for $V$in the above equation.

$K=\frac{1}{2}×0.145\mathrm{kg}×(17\mathrm{m}/\mathrm{s}{)}^2$

$=20.95J$

Hence the kinetic energy is,$20.95J$

The kinetic energy of the body is the energy that the body holds due to motion. This is the work done to the body to accelerate from a standstill to a constant speed.

From the result of part (a) and part (b) the rest energy and kinetic energy is expressed as,

$E_0=1.305×{10}^{16}\mathrm{J}$

$K=20.95\mathrm{J}$

Here , hence kinetic energy is much smaller than the rest energy.