Taking an Integral with respect to time to find the work done on a rocket.

Question

I was given a rocket problem for my Calculus 2 class and want to know if it's possible to take an integral of time to find the correct solution. I think it will be easier to explain if I show the problem and expected solution first.

Here is the problem:
A 0.4 kg rocket is loaded with 0.75kg of rocket fuel. After launch, the rocket rises at a constant rate of 4m/s but the rocket fuel is dissipated at a rate of 0.15kg/s. Find the work done in propelling the rocket 20m above the ground.

This is how I found the solution which I know to be the expected path and correct final value:

Rocket:

$m=0.4kg$
$g=9.8m/s^2$
$h=20m$
$$W = mgh =(0.4)(9.8)(20) = 78.4 J$$

Fuel:

$m = 0.75kg$
$Δm = 0.15kg$
$g=9.8m/s^2$
$ m(t) = 0.75 - 0.15t $
$ d=r*t $ where $d=x$ and $r = 4$
so $\frac 4x = t$ $$W = \int^{20}_0 m(t)gdx $$

Substituting in our values:

$$W = \int^{20}_0 (0.75-0.15\frac x4)(9.8)dx $$ $$W = \int^{20}_0 (7.35-1.47\frac x4)dx $$ $$W = 7.35x-1.47\frac {x^2}8 |^{20}_0 $$ $$W = 7.35(20)-1.47\frac {20^2}8=73.5J $$

So the total work done on the rocket is $ 78.4 + 73,.5 = 151.9 J$

Back to my initial question: Is it possible to solve this problem with an integral of time?

I tried to find a time integral like so:

$$W = gΔs\int^x_0 (0.75-0.15t)dt $$

Because g and Δs are constants they can go in front of the integral.
I made our lower bound 0 since it's the start of our time interval.
I set our upper bound to x hoping I would get a solution of 5 because we know from $ d=r*t $
that $ t = 5 $ if we substitute our $4m/s$ and $20m$
Substituting in the values:

$$73.6 = (9.8)(20)\int^x_0 (0.75-0.15t)dt $$ $$73.6= 196(0.75t- \frac {0.15} 2 t^2) $$ $$0= 72t- 14.7t^2-73.6 $$

Solving for t returns $ t = 3.44s$ and $t = 1.45s$
Are these values correct? What am I missing?

Answer 1

You can use an integral with respect to time if you'd like, but I would justify it slightly differently: let's start from your original approach:

$$W \equiv \int F \cdot dr = \int_0^{20} mg dx$$

Now we can consider a change of variables: because we have constant velocity $\dot{x} = 4$ and $x(0) = 0,$ we have $x = 4t.$ (this is your $d = rt$) So, using the substitution $t = \frac{x}4, dx = 4 dt$:

$$\int_0^{20} mg dx = \int_0^5 mg (4dt) = 4g \int_0^5 0.75 - 0.15t dt$$

This is where the error in your attempt was: the correct factor to put in front of the integral was your velocity, not the total distance. (a dimensional analysis also would've confirmed that this was a bit off conceptually)

Continuing, we see that this method agrees with your original attempt:

$$(4 \text{m/s})(9.8 \text{m/s}^2)\int_0^5 (0.75-0.15t) \text{kg} \ dt = (4 \text{m/s})(9.8 \text{m/s}^2)\big(0.75(5) - \frac{0.15}{2}(5)^2\big) \text{kg}\cdot\text{s} = \boxed{73.5 \text{J}}$$

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Edit: this method can also be made to account for the rocket in the same step by adjusting the mass function:

$$4g\int_0^5 (1.15 - 0.15t) dt = (4)(9.8)\big(1.15(5) - \frac{0.15}{2}(5)^2\big) = \boxed{151.9 \text{J}}$$