Arrow strike, Avg force, Conservation of energy

[Sarah0001]

Homework Statement:

The arrow is brought to rest in a distance of 5 mm, what is the average
force of the arrow strike?

Relevant Equations:

(1/2 )mv^2 = Fx
F= mΔv/Δt

The arrow is following projectile motion to the target when released from an archer's bow.
v vertical = 10ms^-1 v horizontal = 50 ms^-1 resultant v = √2600
mass of arrow = 20*10^-3
I attempted to use F avg = mΔv/Δt to calcualte the average force where Δt = 5*10^-3 / √2600
u = √2600 v = 0
then plugging these in I get an answer of 10400N twice that of the actual answer.


The solution uses the conversation of energy:
ΔKE = Fx
all of arrows KE is importated to the target, the arrow does work over a distance of 5mm to bring itself to rest, so loss of KE = work done by arrow on the target.

I understand this is true, but Q1) what is wrong with using F avg = mΔv/Δt to calculate the average force of the arrow exerts.Q2 What am I wrongly assuming by using this formula? and Q3)why doesn't it apply here?

 

[PeroK]

I think you simply used the initial velocity, rather than the average velocity, which is a half of that.

This was in your calculation of $\Delta t$.

 

[jbriggs444]

I think you simply used the initial velocity, rather than the average velocity, which is a half of that.

This was in your calculation of $\Delta t$.

Average velocity is half of initial velocity under the additional assumption that the deceleration is uniform. The problem statement mentions an "average force". It is clear, accordingly, that such an assumption is not warranted. This makes it impossible to correctly determine a time interval over which the deceleration takes place. Which, in turn, makes it impossible to determine a time-based average for force. [When "average force" is mentioned, a time-based average is normally assumed]

The problem cannot be answered as it stands. It is improperly posed.

However, one can repair this lapse in one of three ways:

One could assume that the deceleration is approximately uniform, use this to estimate a time interval and calculate an approximate average force.

Alternately, one could decide that a distance-weighted average still counts as an "average" and proceed to exactly calculate average force over distance, likely using an energy argument.

Finally, one could decide that a distance-weighted average will approximately match a time-weighted average for reasonably uniform force patterns and that a distance-weighted average will serve as an approximate answer for the question as posed.

 

[PeroK]

Not again! I just let it slide now.

PS that the average velocity is half the initial velocity is actually the necessary and sufficient condition in these cases - in order to do the same calculation as in the constant force case.