A 0.088 kg arrow is fired horizontally. The bowstring exerts an average force of 80 N on the arrow over .75m?

Answer 1

What are you asking? The kinetic energy of the arrow?

Work = force * distance
Work = 0.75 m * 80 N = 60 J

So the kinetic energy of the arrow will be 60 J.

If you want the speed, of the arrow, you can use the kinetic energy of the arrow found above.

Kinetic energy = 1/2 * m * v^2
v = (2 * 60 J/0.088 kg)^(1/2) = 36.9 m/s

If you want to find out how far the arrow will go, we need to know the height of the arrow initially. I'll just do this symbolically and use h for height

x(t) = 36.9 * t
y(t) = 0.5*-9.81 t^2 + h

Those equations will give you the position of the arrow at time t seconds. Solving y(t) for zero, we see that it hits the ground at 0.451524 * h^(1/2) seconds. So, the arrow travels 36.9 * 0.451524 * h^(1/2) meters.

Answer 2

If the question is to find the velocity on exit from the bow, use conservation of energy:

.5*m*v^2=F*d
.5*0.088*v^2=80*.75

If you want to know how far it will fly before hitting the ground:

The horizontal distance is:

v*t=d

t is found by looking at gravity

1/2*g*t^2=height

where height is the height of the arow on exit.

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