 #1
Sarah0001
 31
 1
 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?
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?