A weather rocket is launched straight up. Seconds into the flight, its velocity is 130m/s and it is accelerating at 19 m/s^2. At this instant, the rocket's mass is 59 kg and it is losing mass at the rate of 0.55 kg/s. What is the net force on the rocket?

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A weather rocket is launched straight up. Seconds into the flight, its velocity is 130m/s and it is accelerating at 19 m/s^2. At this instant, the rocket's mass is 59 kg and it is losing mass at the rate of 0.55 kg/s. What is the net force on the rocket?

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Answer 1 · 2015-09-18T14:06:41

1049,5 N up

Explanation:

Rocket's motion depends on Newton 2 and Newton 3.

At time t, rocket has mass m, moves with velocity $\to v$ $vecv$ ejecting fuel in opposite direction with velocity $\to u$ $vecu$ at a rate $- \frac{d m}{d t}$ $-(dm)/dt$ .

At time $t + d t$ $t+dt$ , velocity is now $\to v + d \to v$ $vecv+dvecv$ , and mass $m - d m$ $m-dm$ .

Therefore momentum at time t is $\to p (t) = m \to v$ $vecp(t)=mvecv$

Momentum at time t + dt is $\to p (t + d t) = {\to p}_{r o c k e t} + {\to p}_{f u e l}$ $vecp(t+dt)=vecp_(rocket)+vecp_(fuel_$
$= (m - d m) (\to v + d \to v) + d m [(\to v + d \to v) - \to u]$ $=(m-dm)(vecv+dvecv)+dm[(vecv+dvecv)-vecu]$
$= m \to v + m d \to v - \to u d m$ $=mvecv+mdvecv-vecudm$

Therefore change in momentum is $d \to p = \to p (t + d t) - \to p (t) = m d \to v - \to u d m$ $dvecp=vecp(t+dt)-vecp(t)=mdvecv-vecudm$

$therefore vecF=(dvecp)/dt=d/dtmvecv = m(dvecv)/(dt)-vecu (dm)/(dt)$

This is known as the Rocket Equation and may be used to solve this problem.

Substituting in we get

$sumvecF=59xx19 - 130xx0,55 = 1049,5 N$ upwards

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A weather rocket is launched straight up. Seconds into the flight, its velocity is 130m/s and it is accelerating at 19 m/s^2. At this instant, the rocket's mass is 59 kg and it is losing mass at the rate of 0.55 kg/s. What is the net force on the rocket?

1 Answer

Explanation:

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