Answer:
a) The translational kinetic energy of the particle at point A is 17.28 joules.
b) The speed of the particle at point B is approximately 5.270 meters per second.
c) The total work done on the particle as it moves from A to B is - 9.78 joules.
Explanation:
Let be this particle a conservative system, that is, that non-conservative forces (i.e. friction, viscosity) are negligible.
a) The translational kinetic energy of the particle ([tex]K[/tex]), measured in joules, is determined by the following formula:
[tex]K = \frac{1}{2}\cdot m \cdot v^{2}[/tex] (1)
Where:
[tex]m[/tex] - Mass, measured in kilograms.
[tex]v[/tex] - Speed, measured in meters per second.
If we know that [tex]m = 0.54\,kg[/tex] and [tex]v = 8\,\frac{m}{s}[/tex], the translational kinetic energy at point A is:
[tex]K = \frac{1}{2}\cdot (0.54\,kg)\cdot \left(8\,\frac{m}{s} \right)^{2}[/tex]
[tex]K = 17.28\,J[/tex]
The translational kinetic energy of the particle at point A is 17.28 joules.
b) The speed of the particle is clear in (1):
[tex]v = \sqrt{\frac{2\cdot K}{m} }[/tex]
If we know that [tex]K = 7.5\,J[/tex] and [tex]m = 0.54\,kg[/tex], then the speed of the particle at point B:
[tex]v = \sqrt{\frac{2\cdot (7.5\,J)}{0.54\,kg} }[/tex]
[tex]v\approx 5.270\,\frac{m}{s}[/tex]
The speed of the particle at point B is approximately 5.270 meters per second.
c) According to the Work-Energy Theorem, the total work done on the particle as it moves from A to B ([tex]W_{A\rightarrow B}[/tex]), measured in joules, is equal to the change in the translational kinetic energy of the particle. That is:
[tex]W_{A\rightarrow B} = K_{B}-K_{A}[/tex] (2)
If we know that [tex]K_{A} = 17.28\,J[/tex] and [tex]K_{B} = 7.50\,J[/tex], then the change in the translational kinetic energy of the particle is:
[tex]W_{A\rightarrow B} = 7.50\,J-17.28\,J[/tex]
[tex]W_{A\rightarrow B} = -9.78\,J[/tex]
The total work done on the particle as it moves from A to B is - 9.78 joules.
