Question

The exhaust velocity of gases with respect to a small rocket mass of 25 kg is 2800 m/s. At what rate the fuel must burn so that it may rise up with an acceleration of 9.8 m/s?

Answer 1

Answer:

$\displaystyle{\dfrac{dm}{dt}=0.175 \ kg/sec \ or \ 175 \ g/sec}$

Explanation:

Given :

Relative velocity ( v ) = 2800 m / sec .

Mass ( M )  = 25 kg .

Acceleration ( a ) = 9.8 m / sec .

We know that Thrust on rocket

F  =  u d m / d t

From Newton second law we have

mg + ma = F

ma = F - mg

Now put the value here we get

25 × 9.8 = 2800 × d m / d t - 25 × 9.8

2800 × d m / d t = 25 × 9.8 + 25 × 9.8

2800 × d m / d t = 2 × 25 × 9.8

$\displaystyle{\dfrac{dm}{dt}=\dfrac{25\times9.8\times2}{2800}}\\\\\displaystyle{\dfrac{dm}{dt}=\dfrac{25\times9.8}{1400}}\\\\\displaystyle{\dfrac{dm}{dt}=\dfrac{25\times0.7}{100}}\\\\\displaystyle{\dfrac{dm}{dt}=\dfrac{0.7}{4}}\\\\\displaystyle{\dfrac{dm}{dt}=0.175 \ kg \ or \ 175g}$

Thus we get  rate the fuel is 175 g / sec .

Answer 2

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