If a rocket with a mass of $700 kg$ vertically accelerates at a rate of $3/2 m/s^2$ , how much power will the rocket have to exert to maintain its acceleration at 4 seconds?

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Answer 1 · 2017-01-29T14:47:35

P=6.3 KW

see

MASS=700 kg
Acceleration= $3/2m/s^2$

THEREFORE force exerted is $m*a=700*3/2=1050N$

Now

$a=(v-u)/t$ and we have U ,i.e initial velocity =0 as it started from rest and ofcourse every rockets before launching is at rest,lol.

so by this we can get $v$ and $t$ we are given 4 seconds

$3/2=v/4$
$v=6 m/s$

therefore

$E=F*Displacement$
dividing both sides of the equation by $t$

$E/t=F*(displacement)/t$

And by definition power is energy per unit time and velocity is displacement per unit time.so applying these

$p=F*v$
$P=1050*6$ Watt

$P=6300 W$
$P= 6.3 KW$

If a rocket with a mass of 700 kg700 kg vertically accelerates at a rate of 3/2 m/s^2 3/2 m/s^2, how much power will the rocket have to exert to maintain its acceleration at 4 seconds?