Answer:
a) [tex]v_{f}=\sqrt{v_{o}^{2}+2\cdot a \cdot d}[/tex], b) [tex]v_{f} = 2.877 \times 10^{7} \,\frac{m}{s}[/tex], c) [tex]\Delta K = m \cdot a \cdot d[/tex], d) [tex]\Delta K = F \cdot d[/tex], e) [tex]\Delta K = 2.104 \times 10^{-13}\,J[/tex]
Explanation:
a) The final speed of the proton after travelling a distance is given by this formula:
[tex]v_{f}^{2} = v_{o}^{2} + 2 \cdot a \cdot d[/tex]
[tex]v_{f}=\sqrt{v_{o}^{2}+2\cdot a \cdot d}[/tex]
b) The numerical value is:
[tex]v_{f} = \sqrt{(2.4\times 10^{7}\,\frac{m}{s})^{2}+2\cdot (3.6 \times 10^{15}\,\frac{m}{s^{2}})\cdot (0.035 m) }[/tex]
[tex]v_{f} = 2.877 \times 10^{7} \,\frac{m}{s}[/tex]
c) The change of kinetic energy is modelled after the Work-Energy Theorem:
[tex]\Delta K = \frac{1}{2}\cdot m \cdot (v_{f}^{2}-v_{o}^2)[/tex]
[tex]\Delta K = m \cdot a \cdot d[/tex]
d) The previous expression is re-written as follows:
[tex]\Delta K = F \cdot d[/tex]
e) The numerical value is:
[tex]\Delta K = (1.67 \times 10^{-27} \, kg)\cdot (3.6 \times 10^{15}\,\frac{m}{s^{2}} )\cdot (0.035\,m)[/tex]
[tex]\Delta K = 2.104 \times 10^{-13}\,J[/tex]
