Question

If a rocket with a mass of `900` `"tons"` vertically accelerates at a rate of `3` `"m/s"^2`, how much power will the rocket have to exert to maintain its acceleration at 2 seconds?

Answer 1

`P_"thrust" = 3.46xx10^7` `"W"`

(if given mass is in metric tons)

Explanation:

I'll assume it is `900` metric tons...

We're asked to find the power output the rocket must have so that it continues this acceleration for `2` `"s"`.

Let's first make the conversion from (metric) tons to newtons:

`900cancel("tonnes")((9807color(white)(l)"N")/(1cancel("tonne"))) = color(red)(ul(8.826xx10^6color(white)(l)"N"`

Its mass is thus

`m = w/g = color(red)(8.826xx10^6color(white)(l)"N")/(9.81color(white)(l)"m/s"^2) = ul(9.00xx10^5color(white)(l)"kg"`

The net force on the rocket gives it an acceleration of `3` `"m/s"^2` upward, so the net vertical force on the rocket is

`sumF_y = (9.00xx10^5color(white)(l)"kg")(3color(white)(l)"m/s"^2) = color(orange)(ul(2.70xx10^6color(white)(l)"N"`

So the thrust force exerted by the rocket is

`sumF_y = F_"thrust" - w`

`F_"thrust" = overbrace(color(orange)(2.70xx10^6color(white)(l)"N"))^"net force" + overbrace(color(red)(8.826xx10^6color(white)(l)"N"))^"weight" = color(green)(ul(1.1526xx10^7color(white)(l)"N"`

`---------------------`

We can now find the work done by the thrust force using the equation

`ul(W_"thrust" = F_"thrust"·s)color(white)(aa)` (one dimensional)

We can find the displacement `s` using the kinematics equation

`ul(s = v_0t + 1/2at^2`

where

• `v_0` is the initial velocity, which is `0` assuming it started from a state of rest

• `t` is the time (`2` `"s"`)

• `a` is the constant acceleration (`3` `"m/s"^2`)

Plugging in known values, we have

`s = 0 +1/2(3color(white)(l)"m/s"^2)(2color(white)(l)"s")^2`

`s = ul(6color(white)(l)"m"`

The work done by the thrust force is thus

`W_"thrust" = (color(green)(1.1526xx10^7color(white)(l)"N"))(6color(white)(l)"m") = color(purple)(ul(6.92xx10^7color(white)(l)"J"`

Now, we can find the power exerted by the thrust force using the equation

`P_"thrust" = (W_"thrust")/t`

where

• `W_"thrust"` is the work done by the thrust force (`color(purple)(6.92xx10^7color(white)(l)"J"))`

• `t` is the time maintaining that work (`2` `"s"`)

Thus, we have

`P_"thrust" = (color(purple)(6.92xx10^7color(white)(l)"J"))/(2color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "3.46xx10^7color(white)(l)"W"" ")|)`

The power the rocket must exert is thus `color(blue)(3.46xx10^7` `color(blue)("watts"`.

You can redo these conversions if you meant a different type of ton, and the results are

• short tons: `color(blue)(3.14xx10^7color(white)(l)"W"`

• long tons: `color(blue)(3.51xx10^7color(white)(l)"W"`

Answer 2

We are required to find power of rocket.
We know that Power `=` Rate of doing work

Explanation:

It is assumed that mass of rocket does not change in the stated 2 seconds. Also air friction is ignored.

Rocket is required to generate power to

1. Overcome force due to gravity
2. maintain acceleration at `3ms^-2` for `2s`.

Total upwards acceleration required `a_t=3+9.81=12.81ms^-2`

Using Newton's Second Law of Motion
`F=ma`
Force exerted by rocket
`F=9xx10^5xx12.81=1.15xx10^7N`

Distance moved by rocket in `2s` is found with the help kinematic equation

`s=ut+1/2at^2`

Inserting given quantities we get
`h=0xx2+1/2xx3xx2^2=6m`

Work done by the force
`W=vecFcdotvecS`
`W=1.15xx10^7xx6=6.90xx10^7J`

Power of Rocket `=W/s=(6.9xx10^7)/2=3.45xx10^7W`