an arrow of mass .04 kg is pulled back 10 cm in the bow, having a force of 25.2 N. It is released 1.5m from the ground...how far will the arrow fly?
(all of the energy from pulling it back converts into the flying arrow)
This problem is missing information.
If the pull force is constant:
pulled 0.10 meters * 25.2 = 2.52 Joules
.5 m u^2 = 2.52 Joules
solve for u
however if the bow behaves like a spring
then average force = 1/2 * max force
and we get 1.26 Joules
You do not say which
If you fire horizontally
solve vertical problem to find time in air
(1/2) g t^2 = 1.5 meters
solve for t
then
horizontal distance = u t
If you fire up at 45 degrees (max range) it is a little more complicated but again solve the vertical problem for time, then do the horizontal problem for distance.
To find the distance the arrow will fly, we can use the principle of conservation of energy. The initial potential energy stored in the bow when the arrow is pulled back will be converted into kinetic energy when the arrow is released. Assuming there is no air resistance and all the energy is efficiently transferred to the arrow, we can equate the initial potential energy to the final kinetic energy.
First, let's calculate the initial potential energy stored in the bow. The potential energy (PE) is given by the equation PE = mgh, where m is the mass of the arrow, g is the acceleration due to gravity, and h is the height from which the arrow is released.
Given:
Mass of the arrow (m) = 0.04 kg
Height from which the arrow is released (h) = 1.5 m
Acceleration due to gravity (g) = 9.8 m/s^2
Initial potential energy (PE_initial) = m * g * h
PE_initial = (0.04 kg) * (9.8 m/s^2) * (1.5 m)
PE_initial = 0.784 J
Next, let's calculate the final kinetic energy of the arrow. 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 the velocity of the arrow.
Since we assume all the potential energy is converted into kinetic energy, we can set PE_initial = KE_final.
0.784 J = (1/2) * (0.04 kg) * v^2
Now, solve for v:
v^2 = (2 * 0.784 J) / (0.04 kg)
v^2 = 39.2 J/kg
v = sqrt(39.2 J/kg)
v ≈ 6.26 m/s
Finally, to find the distance the arrow will fly, we can use the equation of motion: d = v * t, where d is the distance, v is the velocity, and t is the time of flight.
We need to find the time of flight. The vertical motion of the arrow can be treated as free-fall motion, where the initial vertical velocity is zero and the final vertical displacement is -1.5 m (since it will fall back to the ground).
Using the equation h = (1/2) * g * t^2, where h is the vertical displacement and t is the time of flight.
-1.5 m = (1/2) * (9.8 m/s^2) * t^2
-1.5 m = 4.9 m/s^2 * t^2
Solve for t:
t^2 = (-1.5 m) / (4.9 m/s^2)
t^2 = -0.3061 s^2
(taking the positive value since time cannot be negative)
t ≈ sqrt(0.3061 s^2)
t ≈ 0.553 s
Now that we have the time of flight, we can calculate the horizontal distance traveled by the arrow.
Distance (d) = v * t
d = (6.26 m/s) * (0.553 s)
d ≈ 3.46 m
Therefore, the arrow will fly approximately 3.46 meters.