Answer:
Givens:
- [tex]m=0.250 \ kg[/tex]
- The different speeds: [tex]v_{1}=2 \ m/s ;v_{2}=3; m/s; v_{3}=4 \ m/s; v_{4}=5 \ m/s; v_{5}=6 \ m/s[/tex]
So, for each speed we will calculate its kinetic energy using its definition:
[tex]K=\frac{1}{2}mv^{2}[/tex]
For [tex]v_{1}=2 \ m/s[/tex]:
[tex]K_{1}=\frac{1}{2}(0.250 \ kg) (2 \ m/s)^{2}=0.5 \ J[/tex]
For v_{2}=3; m/s:
[tex]K_{2}=\frac{1}{2}(0.250 \ kg) (3 \ m/s)^{2}=1.125 \ J[/tex]
For v_{3}=4 \ m/s:
[tex]K_{3}=\frac{1}{2}(0.250 \ kg) (4 \ m/s)^{2}=2 \ J[/tex]
For v_{4}=5 \ m/s:
[tex]K_{4}=\frac{1}{2}(0.250 \ kg) (5 \ m/s)^{2}=3.125 \ J[/tex]
For v_{5}=6 \ m/s:
[tex]K_{5}=\frac{1}{2}(0.250 \ kg) (6 \ m/s)^{2}=4.5 \ J[/tex]
So, there you have it, each kinetic energy for each speed. The only procedure we did was to replace given values and solve basic operations, that's it.
