A 88g arrow is shot from a bow whose string exerts an average force of 110n on the arrow over a distance of 78cm.What is the speed of the arrow as it leaves the string?
Thanks , alot!
To find the speed of the arrow as it leaves the string, we can use the work-energy principle. The work done on an object is equal to the change in its kinetic energy.
The work done (W) is given by the product of the force (F) and the distance (d) over which the force is applied: W = F * d
In this case, the force exerted by the bowstring is 110N and the distance over which it is applied is 78cm (or 0.78m). Therefore, the work done on the arrow is: W = 110N * 0.78m = 85.8 joules (J).
The work done on the arrow is equal to the change in its kinetic energy. The kinetic energy (KE) is given by the equation: KE = (1/2) * m * v^2
Where m is the mass of the arrow and v is its velocity.
To find the velocity, we need to rearrange the equation to solve for v:
v = sqrt((2 * KE) / m)
Substituting the known values: KE = 85.8J and m = 0.088kg (since 88g = 0.088kg), we can calculate the speed of the arrow as it leaves the string:
v = sqrt((2 * 85.8J) / 0.088kg) = 46.59 m/s
Therefore, the speed of the arrow as it leaves the string is approximately 46.59 m/s.
1/2 m v^2=force*distance.
work in SI units, kg, m, N