Question

If a rocket with a mass of `8000 " tons"` vertically accelerates at a rate of `4 m/s^2`, how much power will the rocket have to exert to maintain its acceleration at 3 seconds?

Answer 1

Power is defined as the rate of doing work. As such Power `P` is given as

`P=vecFcdot vecv`

To maintain acceleration of rocket at a rate of `a=4\ ms^-2`, rocket must over come force of gravity acting downwards. As such using Newton's Second Law of motion we get

`F_uarr=m(a+g)`

Inserting given values we get

`F_uarr=8000xx1000(4+9.81)=1.1048xx10^8\ N`

Assuming rocket starts from rest and acceleration is constant, velocity after `3\ s` is found using kinematic expression

`u=u+at`
`v_uarr=0+4xx3=12\ ms^-1`

We see that `cos theta` in the cross product becomes `=1` as force and velocity are in the same direction.

`:.P=(1.1048xx10^8)xx12=1.32xx10^9\ W`, rounded to two decimal places.