Answer:
A) 1050[J]
Explanation:
First, we must calculate the initial kinetic energy when the cat moves at 2 [m/s].
[tex]E_{1} =0.5*m*v_{o}^{2} \\where:\\m = mass = 15[kg]\\v_{o} = 2[m/s]\\\\therefore\\E_{1} =0.5*15*(2)^{2} \\E_{1}= 30[J][/tex]
Then we can calculate the final velocity with the acceleration and time data that was given.
[tex]v_{f} = v_{o}+a*t\\ v_{f} =2+(2*5)\\v_{f} = 12[m/s][/tex]
And we can calculate the kinetic energy as follows:
[tex]E_{2} =0.5*m*v_{f}^{2} \\E_{2} =0.5*15*(12)^{2} \\E_{2} = 1080[J][/tex]
And now we can calculate the difference in kinetic energy by simply subtracting the values calculated above.
DE = 1080 - 30 = 1050[J]
