- #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?