Question about forces over time

[Mr Davis 97]

I am a little confused about forces when time is considered. I understand the nature of forces, ie that they are a push or a pull and that they induce a change in momentum and thus an acceleration. However, my question is, how could we solve such a problem as "If a force of 2 N is applied to a 4 kg block for 6 seconds, what is the resulting acceleration?" When solving force problems, we usually always refer to Newton's second law, F = ma, but I do not see how this equation helps solving problems that involve the application of force over a time interval.

As a side note, if given a problem such as "If a force of 4 N is applied to a 2 kg box, what is the resulting acceleration?" Is 2 m/s^2 really the correct answer? What if we apply the 4 N for 0.5 s rather than 0.1 s? How does F = ma take this into account?

 

[A.T.]

F = ma, but I do not see how this equation helps solving problems that involve the application of force over a time interval.

It doesn't. It relates instantaneous values at some time point.

However, my question is, how could we solve such a problem as "If a force of 2 N is applied to a 4 kg block for 6 seconds, what is the resulting acceleration?

The acceleration is constant 0.5m/s^2 during those 6s. You can use the time to find the change in velocity and position.

 

[dean barry]

If you consider Newtons f = m * a
But a ( acceleration ) = velocity change (vc) / time
So you get:
f = m * ( vc / t )
vc is what you want

 

[CWatters]

I am a little confused about forces when time is considered. I understand the nature of forces, ie that they are a push or a pull and that they induce a change in momentum and thus an acceleration. However, my question is, how could we solve such a problem as "If a force of 2 N is applied to a 4 kg block for 6 seconds, what is the resulting acceleration?" When solving force problems, we usually always refer to Newton's second law, F = ma, but I do not see how this equation helps solving problems that involve the application of force over a time interval.

You only need a=f/m to calculate the resulting acceleration. However part 2 might ask "How fast is it going after 6 seconds?" or "how far has it traveled after 10 seconds"? For that you need different equations. In my school days we learned the SUVAT equations of motion which are mentioned in here..

http://en.wikipedia.org/wiki/Equations_of_motion#Uniform_acceleration

They apply to situations where the acceleration is uniform (constant).

 

[CWatters]

As a side note, if given a problem such as "If a force of 4 N is applied to a 2 kg box, what is the resulting acceleration?" Is 2 m/s^2 really the correct answer?

Yes.

What if we apply the 4 N for 0.5 s rather than 0.1 s? How does F = ma take this into account?

The acceleration will be 2m/s^2 in both cases. However it only continues to accelerate for as long as the force is applied. If the force stops the object will stop accelerating and will then coast along at whatever velocity it had achieved when the force stopped. One of the SUVAT equations is...

V = U + at
where
V = final velocity
U = initial velocity (zero)
a = acceleration
t = time

So if the force is applied for 0.1 seconds the velocity V is...

V = 0 + 2*0.1
= 0.2m/s

If the force is applied for 0.5 seconds the final velocity is

V = 0 + 2*0.5
= 1m/s

 

[dean barry]

Its interesting to note that the impulse ( ( constant ) force * time ) = the momentum change.