An arrow, starting from rest, leaves the bow with a speed of 23.5 m/s. If the average force exerted on the arrow by the bow were doubled, all else remaining the same, with what speed would the arrow leave the bow?
2006-10-06 19:02:51 · 2 answers · asked by Alan l 1 in Education & Reference ➔ Homework Help
Double the speed(47 m/s)
2006-10-06 19:09:58 · answer #1 · answered by Anonymous · 0⤊ 0⤋
The arrow must reach its speed by acceleration from the force of the bow. The acceleration is given by a = f/m, where m is the mass of the arrow. If you assume the distance over which the arrow is accelerated is the same in both cases, the relation between distance and acceleration is given by s = .5*a*t^2. The velocity at which the arrow leaves the bow is v = a*t. From the first equation, t = â(2*s/a). Put this into the equation for velocity to get v = a*â(2*s/a), or v = â(2*a*s). Now a = f/m, so v = (â2*s*f/m). Since s and m are constant, the velocity goes as the square root of the force. Thus the velocity would be â2 * 23.5 for twice the force.
2006-10-06 20:19:10 · answer #2 · answered by gp4rts 7 · 0⤊ 0⤋