Question

If a rocket with a mass of 2900 tons vertically accelerates at a rate of `2/9 m/s^2`, how much power will the rocket have to exert to maintain its acceleration at 6 seconds?

Answer 1

First work out what force the rocket is exerting to accelerate. Then use: Power = Force x Velocity.

The rocket's power is approximately 35.9 MW after 6 seconds.

Explanation:

The first step is to find what resultant upward force needs to be exerted by the rocket to make it accelerate upwards at `2/9 m s^-2`, which we shall denote as `a_r`. The resultant force `F` is equal to the upward force provided by the rocket's motor, `F_r`, minus the force of gravity `F_g`

Resultant upward force = upward forces - downward forces

`F = F_r - F_g`
Rearrange to make `F_r` the subject:
`F_r = F + F_g`

Newton's 2nd law tells us that:
F = m a
The question tells us the upward acceleration is: `a_r=2/9 m s^-2`, which is provided by the resultant force, so:

`F = m_r times a_r`
where `m_r` is the rocket's mass.

`F_g = m_r times text(acceleration due to gravity) = m_r times g`
Now:
`F_r =m_r times a_r + m_r times g = m_r(a_r + g)`

With an expression for the force, we can now address the power:

Power = Force x Velocity
Velocity = acceleration x time
Power = `m_r(a_r + g) times (a_r times t)`

I see you're from the USA, so we'll use US tons (aka short tons).

`m_r = 2900 text( tons) ~= 2900 text( tons) times 907 text( kg per ton = 2630300 kg`

`a_r = 2/9 m s^-2`
`g ~= 10 m s^-2` (use a more accurate value if you want)
t = 6 seconds

Finally, putting it all together:
Power `~= (2630300 kg)(92/9 m s^-2) times 2/9 m s^-2 times 6s`
Power `~= (241987600/9 N) times 12/9 m s^-1`
Power `~= 35850015 J s^-1 = 35850015` watts

Thus the rocket's power is approximately 35.9 MW after 6 seconds.