Question

The point of this question is to compare rest energy and kinetic energy at low speeds. A baseball is moving at a speed of $\mathbf{17}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. Its mass is $\mathbf{145}\mathbf{}\mathit{g}$($\mathbf{0}\mathbf{.}\mathbf{145}\mathbf{}\mathit{k}\mathit{g}$). (a) What is its rest energy? (b) Is it okay to calculate its kinetic energy using the expressionrole="math" localid="1657713286046" $\frac{\mathbf{1}}{\mathbf{2}}\mathit{m}{\mathit{v}}^{\mathbf{2}}$? (c) What is its kinetic energy? (d) Which is true? A. the kinetic energy is approximately equal to the rest energy. B. the kinetic energy is much bigger than the rest energy. C. the kinetic energy is much smaller than the rest energy.

Short answer

(a) The rest energy is, $1.305\times {10}^{\text{'}8}\mathrm{J}$

(b) The kinetic energy is, $20.95\mathrm{J}$

(c) 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

(d) The kinetic energy is much smaller than the rest energy

Step-by-step solution

Identification of the given data

The given data can be listed below as,

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

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

Significance of rest energy

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.

(a) Determination of the rest energy

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 $3\times {10}^8\mathrm{m}/\mathrm{s}$for $c$in the equation (i).

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

$=1.305\times {10}^{16}\mathrm{J}$

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

(b) Determination of the kinetic energy

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.

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

$=20.95J$

Hence the kinetic energy is,$20.95J$

(c) Determination of the kinetic energy

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.

(d) Determination of the kinetic energy is equal to the rest energy or greater than the rest energy

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

$E_0=1.305\times {10}^{16}\mathrm{J}$

$K=20.95\mathrm{J}$

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