Question

What is the kinetic energy of a proton that is traveling at a speed of $\mathbf{3725}\mathbf{\hspace{0.33em}}\mathbf{m}\mathbf{/}\mathbf{s}$ ?

Short answer

The kinetic energy of a proton is $1.16\times {10}^{-20}\hspace{0.33em}\mathrm{J}$.

Step-by-step solution

Identification of the given data

The given data can be listed below as,

• The travelling speed of a proton is, $\mathrm{v}=3725\hspace{0.33em}\mathrm{m}/\mathrm{s}$.

Significance of kinetic energy

The kinetic energy of a body mainly depends on two physical quantities: speed and mass of the body. If the speed of a heavy (large mass) body is more, then the body consists of large kinetic energy.

Determination the kinetic energy of a proton 

The relation of kinetic energy of a proton is expressed as,

$(\mathrm{KE}{)}_{\mathrm{p}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}{\mathrm{v}}^2$

Here,$(KE{)}_p$is the kinetic energy of a proton, and $m_p$is the mass of a proton whose value is $1.67\times {10}^{-27}\hspace{0.33em}\mathrm{kg}$.

Substitute all the known values in the above equation.

$$\begin{gathered} (\mathrm{KE}{)}_{\mathrm{p}}=\frac{1}{2}\left(1.67\times {10}^{-27}\mathrm{kg}\right){\left(3725\mathrm{m}/\mathrm{s}\right)}^2 \\ \approx 1.16\times {10}^{-20}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2 \\ \approx \left(1.16\times {10}^{-20}\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2\right)\times \left(\frac{1\mathrm{J}}{1\mathrm{kg}.{\mathrm{m}}^2/{\mathrm{s}}^2}\right) \\ \approx 1.16\times {10}^{-20}\mathrm{J} \end{gathered}$$

Thus, the kinetic energy of the proton is $1.16\times {10}^{-20}\hspace{0.33em}\mathrm{J}$.