Question

If a rocket with a mass of 3500 tons vertically accelerates at a rate of `1/5 m/s^2`, how much power will the rocket have to exert to maintain its acceleration at 15 seconds?

Answer 1

`P_"thrust" = 5.26xx10^7` `"W"`

Explanation:

I'll assume the given mass is in metric tons.

We're asked to find the necessary power output the rocket must have so that it continues an acceleration for a certain amount of time.

Let's first find the mass of the rocket in kilograms:

`m = 3500cancel("t")((10^3color(white)(l)"kg")/(1cancel("t"))) = ul(3.5xx10^6color(white)(l)"kg"`

Its weight in newtons is thus

`w = mg = (3.5xx10^6color(white)(l)"kg")(9.81color(white)(l)"m/s"^2) = ul(34335000color(white)(l)"N"`

We're given that the acceleration of the rocket is

`a = 0.2` `"m/s"^2`

And so by Newton's second law, the net force acting on the rocket is

`sumF = ma = (3.5xx10^6color(white)(l)"kg")(0.2color(white)(l)"m/s"^2) = ul(7xx10^5color(white)(l)"N"`

The only two forces acting on the rocket are

• its weight (acting downward)

• the thrust force originating from the rocket (acting upward)

We can find the thrust force by the equation for the net force:

`sumF = F_"thrust" - w`

`color(red)(F_"thrust") = sumF + w = 7xx10^5color(white)(l)"N" + 34335000color(white)(l)"N" = color(red)(ul(35035000color(white)(l)"N"`

Now that we know the thrust force, we can begin to calculate the work done by it:

`ul(W_"thrust" = F_"thrust" · s`

To find the displacement `s`, we can use a constant-acceleration equation:

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

The rocket started from rest, the initial velocity `v_0` is `0`, leaving us with

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

We're given

• `a = 0.2color(white)(l)"m/s"^2`

• `t = 15color(white)(l)"s"`

So

`color(green)(s) = 1/2(0.2color(white)(l)"m/s"^2)(15color(white)(l)"s")^2 = color(green)(ul(22.5color(white)(l)"m"`

The work done by the thrust force is thus

`W_"thrust" = color(red)(35035000color(white)(l)"N") * color(green)(22.5color(white)(l)"m") = color(purple)(ul(7.883xx10^8color(white)(l)"J"`

The power output done by the thrust is given by

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

So

`color(blue)(P_"thrust") = (color(purple)(7.883xx10^8color(white)(l)"J"))/(15color(white)(l)"s") = color(blue)(ulbar(|stackrel(" ")(" "5.26xx10^7color(white)(l)"W"" ")|)`

To maintain this acceleration for `15` `"s"`, the rocket's power output must be `color(blue)(5.26xx10^7color(white)(l)"watts"`.