Calculating Projectile Motion: How Changing Force Affects Arrow Speed

[EasyQuestion]

An arrow, starting from rest, leaves the bow with a speed of 25.0 m/s . If the average force exerted on the arrow by the bow were double, all else remaining the same,with what speed would the arrow leave the bow

 

[Hakins90]

First case -

Fd = 1/2mv^2
v^2 = 2Fd/mv
v = sqrt(2Fd/mv)
25 = sqrt(2Fd/mv) - Equation (1)

Second case
(2F)d = 1/2mv^2
v^2 = 4Fd/mv
v = sqrt(4Fd/mv)
= sqrt(2)*sqrt(2Fd/mv)
= sqrt(2) * 25 (from (1))
v = 35.4 ms-1(3sf)

I think :D

 

[EasyQuestion]

Fd = 1/2mv^2?

 

[EasyQuestion]

isnt it d=1/2av^2
d=1/2(f/m)v^2
md=1/2(f)v^2?

 

[Hakins90]

Hmm I am not familiar with those equations...

Work done is Force x Distance.

Since all the work goes into the kinetic energy of the arrow -

Work Done = Kinetic Energy
Fd = 1/2*mv^2

Yeah?

 

[EasyQuestion]

hey, i never learned work done equations, we only used the big five kinematic equations