# Does constant velocity means no force applied?

If there is a body moving with a constant velocity of $$1000m/s$$ and it’s mass =$$10kg$$.

Then , how is it possible that the force applied by this body$$= 0 N$$since it will definitely make an object in front of it move.

$$F_{applied}$$ $$= 10kg * acc(0) = 0N.$$

EDIT: • "force applied by this body or on this body" ??? A good diagram will make the question clear.
– AJN
Jun 21, 2021 at 12:45
• When two bodies collide, there are (equal and opposite) forces acting on both of them. Jun 21, 2021 at 14:39

With constant velocity, the force on the moving object is constant at the interval from the initial position to the final position, so the net change in force is zero. The graph below explains how is so. Note the object needs to accelerate from point 1 to point 2 to reach the velocity of 1000 m/s, thus the net change in force is 100N. The velocity from point 2 to point 3 is constant, a = 0, thus the net force change is 0N, but the force at point 3 = 100 + 0 = 100N (assume zero friction and air resistance, nothing to stop the motion). The force is mostly realized when the object travels with the same velocity from point 3 and hitting the barrier wall at point 4. Depending on the deformation characteristics of the object and the wall, the wall will exert a reactive force on the object to stop the motion, thus, F >= 100N.

• But in your 2nd and 3rd pic of block. There is a constant force on body = 100N which will cause constant acceleration. Therefore , your velocity keeps on increasing. Shouldn’t we take a friction surface with a constant force in this case depending on the situation of friction value or frictionless surfaces with just a force for 1 sec but not constantly applied ? Jun 23, 2021 at 6:54
• Force will not disappear naturally but change form to mechanical energy.
– r13
Jun 23, 2021 at 16:00
• Hey , at t=3. You’re saying that net force on block b/w t=2-3 is 0N. Very impressive. So , acceleration b/w t=2-3 is 0m/s^2. But , I have a counter question. How can Force at t=3 is still 100N. Because for a force to be present , acceleration needs to be present. So , if after t=3 , the car is not accelerating. How is force present at t=3 ? Apr 17 at 9:02
• Even if friction = 0. How can force still be present at t=3 Apr 17 at 9:03
• Conservation of energy, KE = mv^2/2 will not disappear if no friction, it turns into work energy = F*d. The dissipation of energy requires friction to drag down/reduce the speed (deacceleration), or turn the energy into another form - impact/collision to stop the motion at once, at then, v --> 0, and the resulting a is at a peak.
– r13
Apr 17 at 15:17

Accelerating a frictionless body requires force indeed. However keeping a frictionless body at the same velocity requires no action at all. Imagine a rock floating in space. if you kick it, it will float away at a constant velocity, forever. This is exactly the logic behind the famous first law of Newton, which states that if the net force of an object is 0, the velocity is constant.

However, every body on earth is exposed to some kind of friction (that is, every body that is not contained in a vacuum container, levitated in some way). This friction will constantly slow you down. So unless you keep applying force to the body to compensate for said friction, you cannot keep the velocity constant. Which again refers to Newton's law: the net force is zero. $$F_{net} = 0 = -F_{friction}+F_{external}$$

• Mostly okay, except for kicking the rock. By kicking the rock a force is being applied to the rock.
– Fred
Jun 23, 2021 at 8:58

There are many ways to prove that constant velocity means that the resultant force is zero.

## impulse

constant velocity means that from state 1 to state 2 $$\Delta v_{12} = v_{2} -v_{1} =0$$

Therefore, the change in momentum p is also: $$\Delta p_{12} = \Delta m v_{12} = m\cdot v_{2} -m \cdot v_{1} =0$$

However, the change of momentum is called impulse J in physics $$J = \Delta m v_{12} =0$$

And Impulse is also defined as:

$$J = F\delta t$$

Therefore:

$$J = F\delta t = \Delta m v_{12}$$ $$F= \frac{\Delta m v_{12} }{\delta t } =\frac{0}{\delta t} = 0$$

## Newton's second law.

Because v is constant, therefore acceleration is zero.

Therefore:

$$F= m\cdot a= 0$$

## why does it move the block that comes into contact with

The answer to that is the:

• conservation of momentum (and angular momentum)
• conservation of energy.

When the mass hits the object, what happens is that momentum is conserved. So if mass 2 (object) initially was at rest, it will start to move. In order to move, the mass 1 exerts a force $$F_{12}$$ to the second object.

However that means that the momentum of the mass 1 (10 kg) will change as well. Because of that change the net resultant force exerted will be $$F_{21} = - F_{12}$$.

This is the third law of Newton, when two objects interact, they apply forces to each other of equal magnitude and opposite direction.

## Friction

Notice that above, I intentionally write, "resultant force", for the period that the object travels without interacting with mass 2. The reason is that if friction is present, then there is a need for an motive force which is equal and opposite to friction.

• If the motive force is greater than friction, then the object will accelerate.

• If the motive force is less than friction, then the object will decelerate.

In both cases the force is not constant.

I believe the other answers are missing your question.

Well, object with constant velocity (let us call it object 1) is indeed applying force to move the second, the stationary object (let us call it object 2). What force does it apply? The force applied by the object 1 to the object 2 is electrical force. When they get too close their electrons push each other away (they actually always push each other no matter how far, but it's negligible for large distances [most distances are large compared to atomic scales]). The problem is, we can't use this information for modeling your problem, at least not practically. So what we do to model a problem similar to yours is to use the concept of conservation of energy and momentum.

I believe this answers your question, if there is a point I'm missing please notify me.

• Your answer seems interesting but do u have any link to your answer , proof ? Apr 18 at 6:59
• @S.M.T What do you mean a link? Also, this is regular physics: we model problems. Apr 18 at 11:22