Understanding Acceleration and Velocity in Relation to Gravitational Force

AI Thread Summary
It is possible to move an object upwards at a constant velocity, but the net force must be zero, meaning the upward force must equal the downward gravitational force. For a block weighing 20 Newtons, an upward force of 20 Newtons is required to maintain a constant velocity of 5 m/s. Initially, a greater force is needed to accelerate the object upwards before reducing it to 20 Newtons to sustain that speed. Additionally, as the object moves further from the Earth's surface, the gravitational force decreases, which affects the required force to maintain constant velocity. Understanding these dynamics is crucial for accurately applying Newton's laws of motion in real-world scenarios.
cb767
Messages
6
Reaction score
0
I can't believe I don't remember this. Is it possible to move upwards with a constant velocity? For example, if a block weighs 20 Newtons, and I want it to go at a constant velocity of 5 m/s , could I do that with a constant force? For some reason I don't think I can, but maybe I'm just screwing myself up somewhere...
 
Physics news on Phys.org
Force= mass times acceleration. In order that something move with constant velocity it is necessary that the acceleration be 0 and so that the net force be 0. If an object weight 20 N. that means that there is a force of 20 Newtons downward. To balance that and get a net force of 0, you need an upward force of 20 N also. Notice that the "5 m/s" doesn't come into that. In order to move an object weighing 20 N upward at 5 m/s you must first apply a force greater than 20 N upward so there is an acceleration upward. When the speed gets to 5 m/s then you must reduce the upward force to exactly 20 N to keep that constant speed.
 
Incidentally,your case would work out fine,if you were to drag the body with a constant force which would balance the friction force perfectly (i.e.no vertical movement).

Daniel.
 
The amount of force needed to move upwards at constant velocity is the weight of the object.
 
as u go further away from the Earth surface the gravitaional force will continously reduce .so even if u were to ignore the intial extra force applied there would still be an accelleration whose rate would increase the further u went from the earth
 
I have recently been really interested in the derivation of Hamiltons Principle. On my research I found that with the term ##m \cdot \frac{d}{dt} (\frac{dr}{dt} \cdot \delta r) = 0## (1) one may derivate ##\delta \int (T - V) dt = 0## (2). The derivation itself I understood quiet good, but what I don't understand is where the equation (1) came from, because in my research it was just given and not derived from anywhere. Does anybody know where (1) comes from or why from it the...
Back
Top