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.
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Kg m/s N s and N
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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
Now, if you divide the impulse by the time during which the force was applied, you obtain the average force during that time:
(if you know integral calculus, the impulse is the integral of the force over 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/s^{2}. But kg·m/s^{2} is exactly how a newton is defined:1 newton = 1 kg·m/s^{2}
And newtons are a measure of force. So, when you divide the impulse by time, you are calculating the (average) force.

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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
Thanks in advance
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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.
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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
Thanks in advance "
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
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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.
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Re: Kg m/s N s and N
Originally posted by Unregistered View PostIt 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.
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.
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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.
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Re: Kg m/s N s and N
THANK YOU SO Much
Originally posted by Roy Nakatsuka View PostNewtons 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
Now, if you divide the impulse by the time during which the force was applied, you obtain the average force during that time:
(if you know integral calculus, the impulse is the integral of the force over 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/s^{2}. But kg·m/s^{2} is exactly how a newton is defined:1 newton = 1 kg·m/s^{2}
And newtons are a measure of force. So, when you divide the impulse by time, you are calculating the (average) force.
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Re: Converting gravity to Newtons
Originally posted by Unregistered View PostHow do you convert the force of gravity (9.8m/s2) into Newtons?
The "official" value for standard gravity is 9.80665 m/s² (or N/kg). However local gravity varies about ±0.5% from equator to pole at sea level and also varies with altitude. It is common to just assume standard gravity to estimate the forces and rely on the safety margin of a structure. However, if the force is needed to high accuracy, you must determine local gravity. There are formulas that consider latitude and height above sea level.
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Re: Kg m/s N s and N
I have this problem:
While two forces act on it, a particle is to move at the constant velocity = (2.85 m/s)  (3.88 m/s). One of the forces is 1 = (2.45 N) + (5.35 N). What is the other force?
how do I convert m/s to N?
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Re: Kg m/s N s and N
Originally posted by Unregistered View PostQuestion: F = m * a
a = denoted as m/sec2
how does this relate to N/kg?
In the expression N/kg, if you replace newton with kg·m/s², the kg cancel out, leaving acceleration, m/s².
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