A proton that initially is travelling at a speed of $\mathbf{300}\mathbf{m}\mathbf{/}\mathbf{s}$ enters a region where there is an electric field. Under the influence of the electric field the proton slows down and comes to a stop. What is the change in kinetic energy of the proton?

The given data can be listed below as,

• The initial speed of a proton is, ${\mathrm{u}}_{\mathrm{p}}=300\text{\hspace{0.17em}m/s}$.
• The final speed of the proton is, ${\mathrm{v}}_{\mathrm{p}}=0$ (Since the proton slows down and comes to a stop).

In this question, the concept of kinetic energy change of a proton particle will be determined by using the initial and final speed of the proton. If the particle's speed decreases, then the kinetic energy change will be decreased.

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

${\mathrm{\Delta KE}}_{\mathrm{p}}=\frac{1}{2}{\mathrm{m}}_{\mathrm{p}}[{\left({\mathrm{v}}_{\mathrm{p}}\right)}^2-{\left({\mathrm{u}}_{\mathrm{p}}\right)}^2]$

Here, ${\mathrm{\Delta KE}}_{\mathrm{p}}$ is the change in kinetic energy of the proton and ${\mathrm{m}}_{\mathrm{p}}$ represents the mass of a proton whose value is$(1.67\times {10}^{-27}\text{\hspace{0.17em}kg})$ .

Substitute all the known values in the above equation.

$\begin{array}{c}{\mathrm{\Delta KE}}_{\mathrm{p}}=\frac{1}{2}(1.67\times {10}^{-27}\mathrm{kg})[{\left(0\text{\hspace{0.17em}m/s}\right)}^2-{\left(300\text{\hspace{0.17em}m/s}\right)}^2]\\ \approx -7.52\times {10}^{-23}\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}\\ \approx (-7.52\times {10}^{-23}\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}})\times \left(\frac{1\text{\hspace{0.17em}J}}{1\text{\hspace{0.17em}kg}\cdot {\text{m}}^{\text{2}}{\text{/s}}^{\text{2}}}\right)\\ \approx -7.52\times {10}^{-23}\text{\hspace{0.17em}J}\end{array}$

Here, the negative sign indicates that the kinetic energy of the proton decreases as the proton slows down and comes to a stop.

Thus, the change in kinetic energy of the proton is$-7.52\times {10}^{-23}\text{\hspace{0.17em}J}$ .