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#1
09-28-2007, 08:56 AM
 Unregistered Guest Posts: n/a
Kg m/s N s and N

What is the difference between Kg m/s and Newtons when talking about the average force? This is for a problem regarding momentum and impulse, etc.
#2
09-30-2007, 10:20 AM
 Roy Nakatsuka Guest Posts: n/a
Re: Kg m/s N s and N

Newtons are a measure of force. An average force would also be measured in newtons. In contrast, momentum and impulse are measured in kg·m/s.

When a force is applied to an object for a period of time, the force delivers an impulse to the object. An impulse is a change in momentum:
Force x Time = Impulse = Change in momentum
(if you know integral calculus, the impulse is the integral of the force over time)
Now, if you divide the impulse by the time during which the force was applied, you obtain the average force during that time:
Avg force = Impulse ÷ Time
Remember, the units of momentum and impulse are kg·m/s. So, dividing an impulse by time gives you kg·m/s2. But kg·m/s2 is exactly how a newton is defined:
1 newton = 1 kg·m/s2
And newtons are a measure of force. So, when you divide the impulse by time, you are calculating the (average) force.
#3
10-17-2009, 10:40 AM
 Unregistered Guest Posts: n/a
Re: Kg m/s N s and N

Im just wondering I have data in N/s is this the same as kg.m/s or how do i change it to kg.m/s

#4
10-28-2009, 07:22 PM
 Unregistered Guest Posts: n/a
Re: Kg m/s N s and N

looks to me like you still need to find a distance something moved over that time to get power.
#5
05-17-2010, 07:58 PM
 Unregistered Guest Posts: n/a
Re: Kg m/s N s and N

"Im just wondering I have data in N/s is this the same as kg.m/s or how do i change it to kg.m/s

Answer: Doing problems like this requires unit analysis.

Newton=mass(kg)*acceleration due to gravity(m/s^2)=kg*m/sec^2

Time=sec

The only way to relate time and force the way you request is to multiply the two
N*Time=(kg*m/sec^2)*sec=kg*m/sec or N*s
*the sec^2 in the denominator is canceled out to a single sec by the unit of time in the numerator of the multiplier

the unit analysis for N/s is as follows:
N=(kg*m/sec^2)*(1/sec)=kg*m/s^3

Depending on the context of the problem you may or may not be able to use these units.
For example, if you know that a force of 20 N was applied to hold a box in static equilibrium for one minute, then it is known that 1200 N*s or 1200 kg*m/sec is applied to the box. this is the only type of problem I can think of relating the two.

hope this helps
#6
05-23-2010, 03:42 PM
 Unregistered Guest Posts: n/a
Re: Kg m/s N s and N

It seems unlikely that you would have a measurement in N/s. That would be (as stated above) a unit of kg.m/s3, which seems strange.

Is it possible that your unit is actually N.s? (That is, Newtons * seconds). That would in fact be equal to kg.m/s, which is what you wanted.
#7
05-24-2010, 08:41 AM
 JohnS Double Ultimate Supreme Member Join Date: Dec 2007 Location: SE Michigan, USA Posts: 9,218 Rep Power: 18
Re: Kg m/s N s and N

Quote:
 Originally Posted by Unregistered It seems unlikely that you would have a measurement in N/s. That would be (as stated above) a unit of kg.m/s3, which seems strange. Is it possible that your unit is actually N.s? (That is, Newtons * seconds). That would in fact be equal to kg.m/s, which is what you wanted.
F*t, or the integral for a varying force, is impulse, equal to the change in momentum of the body acted upon. It is very common in rocket problems, but applicable to other things.

However, the derivative of force with respect to time is also important, dF/dt = m*da/dt. The derivative of acceleration, da/dt, is known as jerk, and is closely related to "ride quality" in everything from cars to elevators (Amusement park rides may consider high jerk good).

As both have uses, I think we need a better description of the data and its meaning or use from the poster who asked the question.
#8
07-18-2010, 05:29 AM
 raoe Guest Posts: n/a
Re: Kg m/s N s and N

It is possible to have N/s units.

Consider calculating the power required for a motor moving fluid. The fluid has a weight of so many N, is moving up a distance of so many m, and this is happening each s. So, the power required would be in (N/s)*m = W. But what is the name of this intermediate unit in kg m / s3? It seems to have no name.
#9
02-06-2011, 08:28 PM
 Unregistered Guest Posts: n/a
Re: Kg m/s N s and N

THANK YOU SO Much

Quote:
 Originally Posted by Roy Nakatsuka Newtons are a measure of force. An average force would also be measured in newtons. In contrast, momentum and impulse are measured in kg·m/s. When a force is applied to an object for a period of time, the force delivers an impulse to the object. An impulse is a change in momentum:Force x Time = Impulse = Change in momentum (if you know integral calculus, the impulse is the integral of the force over time)Now, if you divide the impulse by the time during which the force was applied, you obtain the average force during that time:Avg force = Impulse ÷ TimeRemember, the units of momentum and impulse are kg·m/s. So, dividing an impulse by time gives you kg·m/s2. But kg·m/s2 is exactly how a newton is defined:1 newton = 1 kg·m/s2And newtons are a measure of force. So, when you divide the impulse by time, you are calculating the (average) force.
#10
03-08-2011, 07:34 AM
 success makurdi Guest Posts: n/a
Re: Kg m/s N s and N

helo guys am realy gr8ful you guys have just saved me

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