Reviewing Kinematics for AP Physics Exams: When to Subtract Gravity?

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In reviewing kinematics for AP Physics, it's clarified that when a rocket accelerates upwards at 5 m/s², this value already accounts for overcoming gravity, so there's no need to subtract 9.8 m/s². The net acceleration of 5 m/s² indicates the rocket has successfully overpowered gravitational pull. When stationary, the forces of gravity and the ground balance each other, resulting in zero acceleration. To calculate thrust, one must consider that the total force must account for both the upward acceleration and the gravitational force acting downwards. Understanding these concepts is crucial for solving rocket motion problems effectively.
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Hello, I'm reviewing kinematics for an upcoming AP physics exam, but I seem to have forgotten a crucial fact. In typical rocket problems (when the rocket blasts off with an initial acceleration for example 5 m/s^2), why don't we subtract 9.8 from 5m/s^2 when using a value for acceleration? When do we know to subtract gravity from an upwards acceleration, or to just simply use the value given for the upwards accel w/o subtracting?
 
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The 5 m/s2 is the acceleration upwards, and this should become clear from the way the value is given. If you want to find the thrust or forces acting on the rocket, you'll have to take into account that the rocket has to fight against gravity.
 
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The value of 5 m/s2 describes the actual motion of the missile. If gravity is the ONLY force acting on an object, then it will accelerate toward the Earth at 9.8 m/s2.

So when it sits on the launchpad, gravity pulls down with a force, but the ground pushes back with the same exact force, resulting in zero acceleration. When the rocket ignites and exhaust exits the engine, the rocket accelerates upward.

Knowing the actual acceleration of 5 m/s2, can you calculate the amount of thrust necessary to maintain that acceleration? (remember that in F=ma, F is the vector sum of all forces acting on an object).
 
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mfb said:
The 5 m/s2 is the acceleration upwards, and this should become clear from the way the value is given. If you want to find the thrust or forces acting on the rocket, you'll have to take into account that the rocket has to fight against gravity.

Oh, so the net acceleration is 5 m/s^2? (As in, the rocket has already overpowered gravity, so 9.8 does not need to be subtracted from 5?)
 
scottdave said:
The value of 5 m/s2 describes the actual motion of the missile. If gravity is the ONLY force acting on an object, then it will accelerate toward the Earth at 9.8 m/s2.

So when it sits on the launchpad, gravity pulls down with a force, but the ground pushes back with the same exact force, resulting in zero acceleration. When the rocket ignites and exhaust exits the engine, the rocket accelerates upward.

Knowing the actual acceleration of 5 m/s2, can you calculate the amount of thrust necessary to maintain that acceleration? (remember that in F=ma, F is the vector sum of all forces acting on an object).

So the thrust will be F=5*mass of the rocket?
 
If you were to subtract 9.8 from 5 you would get a negative number. Perhaps get into the habit of thinking about what the answer means. Eg What would a negative acceleration actually mean? Is it likely a rocket taking off would have a negative acceleration?
 
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belledona said:
So the thrust will be F=5*mass of the rocket?
Let me rephrase that. If acceleration = (net force) / (mass). Net force is vector sum of all forces acting on it. What forces are there and what are the directions?
 
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