Question

A 1500kg space probe is moving at constant velocity of 4200m/sec. In a course change maneuver the rocket engines fire for 5 minutes increasing the probe's velocity to 5000m/sec. What is the probe's change in kinetic energy? A. 3.68x10^6 J. B. 5.52x10^9 J C. 1.1x10^10 J D. not enough info.

Answer 1

B)5.52x10^9 J

Step-by-step explanation:

The equation for calculating kinetic energy is as follows

E=
`(1)/(2) mv^(2)`

so what we have to do is calculate the kinetic energy of the space probe in each state and calculate the difference

E=
`(1)/(2) m.V2^(2)-(1)/(2) m.V1^(2)=E`

E=
`(1)/(2) m(V2^(2) -V1^(2))=E`

E=(1500)(5000^2-4200^2)/2

E=5.52x10^9 J