In the absence of air resistance and friction, what will happen to the velocity of an object going at 20 m/s E?
1 answer:
In the absence of any other forces, the object will continue to move at 20 m/s E.
In fact, Newton's second law states that the resultant of the forces acting on an object is equal to the product between the mass and the acceleration of the object:
therefore, if there are no forces acting on the object, the term on the left is zero, and the acceleration of the object is zero as well. This means that the object will continue its motion with constant speed, and in the same direction.
In fact, Newton's second law states that the resultant of the forces acting on an object is equal to the product between the mass and the acceleration of the object:
therefore, if there are no forces acting on the object, the term on the left is zero, and the acceleration of the object is zero as well. This means that the object will continue its motion with constant speed, and in the same direction.
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The first: alright, first: you draw the person in the elevator, then draw a red arrow, pointing downwards, beginning from his center of mass. This arrow is representing the gravitational force, Fg.
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
let's do it for this case:
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere
You can always calculate this right away, if you know his mass, by multiplying his weight in kg by the gravitational constant
let's do it for this case:
the unit of your fg will be in Newton [N]
so, first step solved, Fg is 637.65N
Fg is a field force by the way, and at the same time, the elevator is pushing up on him with 637.65N, so you draw another arrow pointing upwards, ending at the tip of the downwards arrow.
now let's calculate the force of the elevator
so you draw another arrow which is pointing downwards on him, because the elevator is accelating him upwards, making him heavier
the elevator force in this case is a contact force, because it only comes to existence while the two are touching, while Fg is the same everywhere