newtons second law?

Question

arrow, starting from rest, leaves the bow with a speed of 28.5 m/s. If average force on the arrow by the bow were doubled, all else remaining the same, what speed would the arrow leave the bow?


i'm completely lost by this problem........

Answer 1

F = m*a ------> a = F/m

V^2 = V0^2 - 2*a*(x-x0)

V0 and x0 = 0 (assuming the arrow starts from rest)

Plugging everything in, you get the expression for velocity:

V = sqrt(2*F*x/m)

Substituting 2*F for F, we get:

V(new) = sqrt(4*F*x/m)

Thus, V(new) = sqrt(2) * V

Answer 2

It is better to think in terms of Impulse and Change in Momentum:

Impulse = Change in Momentum Or

F*t = mv -0 where v is the speed after the arrow is released and 0 was the speed before that.
Clearly, if the average force is doubled, everything else remaining the same, v will also be doubled. That is, it will be 57 m/s. Answer.

Answer 3

What clydesdale wrote is correct.

There is an even simpler way, however, which relies on insight rather than computation.

The work done by the bow is what gives the arrow its kinetic energy.
KE(arrow) = integral(F * dx) of the bow
If F doubles, KE doubles
-->: velocity increases by sqrt(2)

Answer 4

Newtons second regulation of stream states that the cost of change in momentum is right now proportional to the stress utilized and the change in momentum is contained in the approach the force utilized.

Answer 5

use F=ma
force is doubled,
2F=2ma
since m remains constant, the acceleratuion is double... according to the equation: v=u+at,
the velocity with which it leaves the bow is doubled.... that is 57m/s

Answer 6

i think velocity will doubble too..
v=u+at
u=0 so first case is 28.5 = at
now force doubled leads to double acc.
hence 2at=v=57m/s