An archer puts a 0.29 kg arrow to the bow- string. An average force of 224.8 N is exerted to draw the string back 1.44 m.
The acceleration of gravity is 9.8 m/s2 .
Assuming no frictional loss, with what speed does the arrow leave the bow?
Answer in units of m/s.
To find the speed at which the arrow leaves the bow, we need to use the principle of conservation of energy.
The potential energy stored in the bow when the string is pulled back is equal to the kinetic energy of the arrow when it is released. Since there is no frictional loss, we can assume that all the potential energy is converted into kinetic energy of the arrow.
First, let's calculate the potential energy stored in the bow. The formula for potential energy is:
Potential Energy = mass * gravity * height
The mass of the arrow is given as 0.29 kg, the acceleration due to gravity is 9.8 m/s^2, and the height is the distance the string is pulled back, which is 1.44 m. Plugging in these values:
Potential Energy = 0.29 kg * 9.8 m/s^2 * 1.44 m
= 4.02 J (Joules)
Next, we can equate the potential energy to the kinetic energy of the arrow. The formula for kinetic energy is:
Kinetic Energy = (1/2) * mass * velocity^2
We want to find the velocity, so we can rearrange the equation to solve for velocity:
velocity = sqrt((2 * Kinetic Energy) / mass)
Since the potential energy is converted entirely into kinetic energy:
Kinetic Energy = Potential Energy
Plugging in the values:
velocity = sqrt((2 * 4.02 J) / 0.29 kg)
= sqrt(27.86 m^2/s^2)
= 5.28 m/s
Therefore, the speed at which the arrow leaves the bow is 5.28 m/s.