What is the net force of a rocket that fired vertically from rest with a net acceleration of 17.5 m/s^2.....?

Question

A model rocket is fired vertically from rest. It has a net acceleration of 17.5 m/s^2. After 1.5s, its fuel is exhausted and its only acceleration is that due to gravity. The rocket has a mass of 87g and you may assume the mass of the fuel to be much less than 87g. What was the net force on the rocket during the first 1.5 s after liftoff? What force was exerted on the rocket by the burning fuel? What was the net force on the rocket after its fuel was spent? The rocket's vertical velocity was zero instantaneously when it was at the top of its trajectory. What were the net force and acceleration on the rocket at this instant?

Answer 1

The net force is related to
F=m*a, where F is the net force. Since a=17.5
F=m*17.5

What may confuse you is if the question asked for the thrust of the rocket. The rocket motor must overcome gravity and achieve an upward acceleration. This means that thrust, T, is related by
T-m*g=m*17.5. This is the answer to the second part of the question of the force of the burning fuel.


After the fuel is spent, the net force is m*g. The equation of motion is
v(t)=v0-g*t^2
and
y(t)=y0+v0*t-.5*g*t^2

At the instant of apogee, the net force is still m*g, and the acceleration is -g. The idea of weightlessness may confuse here. "Weightlessness" is achieved if there is no reactive force on a mass. To experience weightlessness for yourself, jump off a garden wall. During the time you are airborne you experience weightlessness.
When you are standing on the ground, the net force on you is zero. This is because the force of gravity is counteracted by the reaction force at the contact point with the ground. Your acceleration is zero, so net force, F=0.

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Answer 2

a) throughout the time of the 1st a million.7 s after liftoff you suggested that we've an acceleration of sixteen.0 m/s^2 (interior the upward direction needless to say). for this reason from newtons regulation: F = ma on the thank you to say (assuming that the mass of the rocket gasoline is negligible): F = [(84g)/(1000g/kg)]*sixteen.0m/s^2 = a million.344 N [upwards] b) The stress exerted on the rocket by ability of the burning gasoline is an element of the internet stress. subsequently we've gravity working interior the downward direction and the burning of the rocket gasoline propelling the rocket up. for this reason from the internet stress, which we in basic terms found out to be a million.344 N (up) we can say: a million.344 N (up) = stress by using Burning of gasoline (up) - stress of Gravity (down) a million.344 N (up) + stress of Gravity (down) = stress by using Burning of gasoline (up) a million.344 N (up) + (0.84g)/(1000 g/kg)*9.8 m/s^2 = stress by using Burning of gasoline (up) a million.3358 N (up) = stress by using Burning of gasoline (up) c) After the gasoline became spent, there is now in basic terms gravity engaged on the rocket interior the downwards direction. for this reason internet stress subsequently = stress of gravity = 0.80 4 grams * a million/1000 g/kg * 9.8 m/s^2 = 0.008232 N. d) This one i'm no longer incredibly particular on, i will bypass away this area to somebody else. keep in mind nonetheless, all of that's assuming the rocket gasoline weighs zilch.