# Finding Velocity for an object in Translational Equilibrium

1. Oct 1, 2015

### Ocata

Hi,

Suppose a force of 5N is being applied to a 5kg block and the force of friction from the surface on which the block is in motion is 5N as well. The Net Force of the block is 0N, thus the block is not accelerating and is moving at a constant velocity.

The question I have is, how to determine the velocity of the block when it's net force is zero? If the only information I have in this scenario is that the mass of the block is 5kg and the translational equilibrium is when the applied force is 5N, then acceleration, time, and distance have not been specified, so don't think I can use kinematic equations to determine the velocity.

Thank you.

2. Oct 1, 2015

### insightful

The velocity would depend on how much power (force X velocity) is available to pull the block.

3. Oct 1, 2015

### Ocata

Thank you, insightful.

I guess that means it is impossible to determine the velocity based only on values for Net Force and mass of an object.

So starting with a coefficient of kinetic friction between a block and the surface on which it is set, I can calculate the force that I must apply in order to generate a net force of zero and thus set the block into a motion of constant velocity.

For instance, if the block is 5kg and the coefficient of kinetic friction is .1, then [mue(Fn) = Ff] = [mue(mg) = Ff] = [.1(5*10) = 5N = Ff].
Thus, when I apply a force of 5N, the block will have a net force of 0 and thus be traveling with a constant velocity.
From here, I can not determine any information about the velocity of the block until I actually measure the distance that the block travels over a certain amount of time.

If my goal is to determine the velocity of a block of a given mass based on the coefficient of kinetic friction, then I need to make sure I have a stopwatch and tape measure in order to measure the distance the block travels over a certain amount of time.

Then, if my goal was to find velocity, I didn't need to know the coefficient of kinetic friction nor the mass of the block. I just need to know the distance traveled over the time interval traveled.

So, the net force on an object seems to be completely independent from the velocity it travels. For example, if an object is traveling in space, the coefficient of kinetic friction is 0, so the velocity of the object will be constant as long as there is no applied force. If the block is traveling at .001m/s, then as long as 0N of Force is applied, the block will continue traveling at that speed. If the block is traveling at 1000m/s, then as long as 0N of force is applied, the block will continue at that speed. Since there seems to be no specific applied force that generates a specific constant velocity based on the coefficient of kinetic friction, then it seems to make sense that it would not be possible to determine the blocks velocity by simply stating that 0N are being applied to it.

But when a force is being applied to the block when the block is on a table with a specific kinetic friction (.1), then there seems to be only one specific applied force (5N) that generates only 1 specific constant velocity. So it just feels like there is somehow more information from which to determine the speed that the block must be traveling when 5N is being applied to it.

From what I understand, Power = f*v = (f*d)/t , then in order to solve for velocity, I would have to already know distance and time, which seems to defeat the purpose of having to know the force in the first place. In addition, in the power formula, isn't the force variable referring to net force? If so, isn't the net force of the block in equilibrium 0N? Wouldn't that cancel out the entire formula anyhow? For example, [Power = (0N)*v = ((0N)*d)/t] = [0 = 0], no?

4. Oct 1, 2015

### insightful

No, the velocity will vary depending on the power applied.

No, it refers to the force being applied to overcome the friction force.

5. Oct 2, 2015

### Ocata

So for instance, the force I am applying is 5N to obtain equilibrium. And power is calculated by the Applied Force of 5N, not the Net Force of 0N.
However, even if power formula refers to the applied force of 5N, velocity still seems undeterminable because force just seems to cancel out of the equation. And if force cancels out of the equation, then the only way to know the velocity is to know the distance and time the block has traveled. So if a value for distance and time are not given, how is it possible to determine the velocity of the block using the power formula?

Power = $f_{a}*v = \frac{f_{a}*d}{t}$

How would you determine the velocity of a 5kg block that is in translational equilibrium when 5N of force is applied to it? It doesn't seem possible to me based on the Power formula.

6. Oct 2, 2015

### nasu

Well, if you knew the power and the force, you can find the velocity.
The problem is to find the values of these parameters. isn't it?

7. Oct 2, 2015

### insightful

No, power is an independent variable. Let's try this: What if a small tractor with a battery-powered 5 watt electric motor were attached to pull our 5 kg block. If geared to use all 5 watts, what is the velocity of the block (assume 100% efficiency for the tractor)?

8. Oct 4, 2015

### Ocata

First guess:

[Power = $\frac{W}{t}]$ = [Power = $\frac{f_{applied}*d}{t}]$ =

[5 Watts = $\frac{5N*d}{t} ]$ = [5 Watts = $5N(v)$ = $\frac{5W}{5N}=v$ = 1m/s

My other guess would be:

[Power = $\frac{W}{t}]$ = [Power = $\frac{f_{applied}*d}{t}]$ = [Power = $\frac{(mg)*d}{t}]$

[5 Watts = $\frac{(5kg)(10\frac{m}{s^{2}})*d}{t} ]$ = [5 Watts = $\frac{50N*d}{t}$ = [5W = $50N(v)$ = $\frac{5W}{50N}=v$ = .1m/s

I have not begun to learn about power yet, but from the equation and from your advice that force refers to applied force, these are my attempts at finding the velocity of 5 Watts on a 5kg block. Hopefully one is correct, if not, then I look forward to learning about what I'm doing wrong so I can make the necessary corrections.

Thank you

Last edited: Oct 4, 2015
9. Oct 4, 2015

### insightful

Your "first guess" is correct. Good job!

In your "other guess" you used mg, which is the normal force the block exerts on the horizontal surface.

Soon you will not be guessing.

10. Oct 4, 2015

### Ocata

I sure hope so, thank you.