Easy acceleration of a rocket question help plz (1 Viewer)

petergriffin198 · Sep 22, 2015

A rocket and its fuel have a mass of 2.5 x 10^6kg as the engines ifnite for take off. The engines deliver a constant thrust of 5 x 10^7 N. Given that 75% of the rocket is fuel, calculate the final acceleration of the rocket.

keepLooking · Sep 22, 2015

$F = 5 \times 10^7$
$m = 2.5\times10^6(0.25)$

$a = \frac{F}{m}$
$a = \frac{5 \times 10^7}{2.5\times10^6(0.25)}$
$a = 80$

(Unsure)

Fizzy_Cyst · Sep 23, 2015

keepLooking said:
$F = 5 \times 10^7$
$m = 2.5\times10^6(0.25)$

$a = \frac{F}{m}$
$a = \frac{5 \times 10^7}{2.5\times10^6(0.25)}$
$a = 80$

(Unsure)

The net force acting on the rocket is Thrust - Weight, not just thrust.

Mr_Kap · Sep 23, 2015

Fizzy_Cyst said:
The net force acting on the rocket is Thrust - Weight, not just thrust.

So what's the answer to this question. How do i do it?

Drsoccerball · Sep 23, 2015

F=T-mg (By equating vector of weight downwards and thrust force upwards)
ma=T-mg
a= T/m -g

You know thrust, you know g and the value of the "new" m is 75/100 x m(initial) then you have the acceleration.

Fizzy_Cyst · Sep 23, 2015

$T = 5 \times 10^7$
$m = 2.5\times10^6(0.25)$

$a = \frac{T-mg}{m}$
$a = \frac{5 \times 10^7 - (2.5\times10^6(0.25) \times 9.8)}{2.5\times10^6(0.25)}$
$a = 70.2$

Mr_Kap · Sep 23, 2015

Fizzy_Cyst said:
$T = 5 \times 10^7$
$m = 2.5\times10^6(0.25)$

$a = \frac{T-mg}{m}$
$a = \frac{5 \times 10^7 - (2.5\times10^6(0.25) \times 9.8)}{2.5\times10^6(0.25)}$
$a = 70.2$

thanks.

Easy acceleration of a rocket question help plz (1 Viewer)

petergriffin198

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keepLooking

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Fizzy_Cyst

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Mr_Kap

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