Question

A runner whose mass is $\mathbf{60}\mathbf{}\mathit{k}\mathit{g}$runs in the $\mathbf{+}\mathit{x}$direction at a speed of $\mathbf{7}\mathbf{}\mathit{m}\mathbf{/}\mathit{s}$. (a) What is the kinetic energy of the runner? (b) Runner turns around and runs in the$\mathbf{-}\mathit{x}$direction at the same speed. Now what is the kinetic energy of the runner?

Short answer

(a) The kinetic energy in $+x$direction is, $1470\mathrm{J}$

(b) The kinetic energy in $-x$direction is,$1470\mathrm{J}$

Step-by-step solution

Identification of the given data

The given data can be listed below as,

- The mass of runner is, $m=60\mathrm{kg}$.

- The speed of runner in $+x$ direction is, $v_1=7\mathrm{m}/\mathrm{s}$

- The speed of runner in -$x$ direction is, $V_2=-7\mathrm{m}/\mathrm{s}$

Significance of rest energy

The energy equivalent to an object's rest mass is equal to its rest mass multiplied by the square of the speed of light.

(a) Determination of the kinetic energy in +x direction

The relation of kinetic energy is expressed as,

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

(i)

Here $K$is the kinetic energy, $m$is the mass of the runner and $V$is the speed of the runner. Substitute role="math" localid="1657720722953" $60\mathrm{kg}$for $\mathrm{m}$and $7\mathrm{m}/\mathrm{s}$for $v_1$in the equation (i).

$K=\frac{1}{2}\times 60\mathrm{kg}\times (7\mathrm{m}/\mathrm{s}{)}^2$

$=1470J$

Hence the kinetic energy in +x direction is,$1470J$ .

(b) Determination of the kinetic energy in -x direction

The relation of kinetic energy is expressed as,

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

(i)

Here $K$is the kinetic energy, $m$is the mass of the runner and $V$is the speed of the runner.

Substitute $60\mathrm{kg}$for $\mathrm{m}$and $7\mathrm{m}/\mathrm{s}$for $v_1$in the equation (i).

$K=\frac{1}{2}\times 60\mathrm{kg}\times (7\mathrm{m}/\mathrm{s}{)}^2$

$=1470J$

Hence the kinetic energy in -x direction is, $1470J$.

Therefore, the kinetic energy in +x and -x direction is same.