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Say you push an object with 10 N force(just once and let it go) of mass 10kg object in space with zero friction and nothing in the way of its path. What would be the velocity of that object?

I would assume that if you are putting a "constant" force of 10N, then yes theoretically the object should continue accelerating forever right?

But what if you just give an initial push of 10N to the object and let it go so that it floats in space how would you calculate its definite velocity? I am assuming the time in which the object is in the contact with a pushing force will matter is that correct?

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  • $\begingroup$ Collision should help. But one has to know something on the pushing body. $\endgroup$ – Alchimista Mar 25 at 9:14
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Yes, the time will matter.

To calculate the final velocity of an object v ,with an inital velocity u,which is under a constant acceleration of a for a time t. We have the relation $$v=u+at$$ The object would continue to move forever with a constant velocity(v) due to Newton's first law , which states that an object continues to remain in its state (of rest or motion) unless acted upon by a net external force.

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If you don't have your space watch, but did bring a ruler, you could measure how far you pushed it (you pushed it $x$). Then the work you did is

$$ W = F\cdot x $$

where $F=10\,$N. That is equal to the object's kinetic energy:

$$ T = \frac 1 2 mv^2 = Fx $$

Of course, the average speed is $v/2$, so the time it takes to go $x$ is:

$$ t = \frac{2x} v $$

which you can plug in:

$$ \frac 1 2 m v^2 = \frac 1 2 v tF $$

or

$$ v = t\frac F m $$

Since $F=ma$, that's:

$$ v = at $$

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Say you push a mass with 10 N force (just once and let it go) of 10kg object in space with zero friction and nothing in the way of its path. What would be the velocity of that object?

It depends on what you mean by “just once and let go” and the reference frame in which the velocity is calculated. In other words, what was the duration of the force (t) or what was the distance (d) over which the force was applied and with respect to what point in space the measurement is made.

Let’s say the frame of reference is a point in space at the same location as the mass before the force was applied to the mass. We can calculate the velocity in a number of ways, one is to use basic kinematics (given the duration of the application of the force) and the other using the work energy principle (given the distance through which the force acts)

Kinematic approach:

$$F=ma$$ $$a=\frac{F}{m}$$ $$v=at$$ Let’s say the force is applied for 1 second, then for $F=10N$ and $m=10kg$, then $$v=1\frac{m}{s}$$

Work-energy principle:

The work-energy principle states that the change in kinetic energy of an object is equal to the net work done on the object. $$W=Fd$$ $$Fd=KE_{final}-KE_{initial}$$ Since the initial velocity is zero (with respect to the frame of reference chosen) we have $$Fd=\frac{mv^2}{2}$$ Let’s say the force is applied over a distance of 0.5 meter. $$(10)(0.5)=\frac{(10)(v^2)}{2}$$ $$v=1\frac{m}{s}$$

I would assume that if you are putting a "constant" force of 10N, then yes theoretically the object should continue accelerating forever right?

No, because the maximum possible speed is the speed of light. But that’s the subject of special relativity.

But what if you just give an initial push of 10N to the object and let it go so that it floats in space how would you calculate its definite velocity? I am assuming the time in which the object is in the contact with a pushing force will matter is that correct?

I think my answer to the first question answers this as well

Hope this helps.

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