To solve this problem, we can use the concepts of work, energy, and projectile motion.
(a) To find the speed at which the arrow leaves the bow, we can use the principle of conservation of mechanical energy, which states that the initial mechanical energy (potential energy + kinetic energy) of an object is equal to its final mechanical energy, assuming no frictional losses.
The initial mechanical energy when the arrow is drawn back is stored in the potential energy of the bowstring. The final mechanical energy is divided between the potential energy and the kinetic energy of the arrow as it leaves the bow.
Let's calculate the initial potential energy of the bowstring:
Initial potential energy (PEi) = Force x Distance
PEi = 200 N x 1.4 m
PEi = 280 Joules
Since there are no frictional losses, the final mechanical energy of the system will be the same as the initial potential energy:
Final mechanical energy (PEf + KEf) = PEi
KEf = PEi - PEf
The final kinetic energy (KEf) will be given by:
KEf = 0.5 * mass * velocity^2
Given that the mass (m) of the arrow is 0.26 kg, we can rearrange the equation to solve for velocity (v):
v = √(2 * KEf / m)
We can substitute the values into the equation to find the velocity:
v = √(2 * (PEi - PEf) / m)
Now, let's find the final potential energy (PEf) of the arrow by converting the kinetic energy of the bowstring into the potential energy of the arrow. Assuming no other energy losses, the potential energy will be the same as the kinetic energy stored in the bowstring:
PEf = KEi
Given that the kinetic energy (KEi) is equal to 280 Joules, we can substitute the value and solve for velocity:
v = √(2 * (PEi - KEi) / m)
v = √(2 * (280 J - 280 J) / 0.26 kg)
v = 0 m/s
Therefore, the speed at which the arrow leaves the bow is 0 m/s.
(b) If the arrow is shot straight up, we can calculate the maximum height it reaches by using the equations of projectile motion.
The maximum height the arrow reaches can be determined using the equation:
h = (v^2 * sin^2 (θ)) / (2 * g)
Where:
- h is the maximum height reached by the arrow
- v is the initial velocity of the arrow (which is 0 m/s, as calculated above)
- θ is the angle at which the arrow is shot (straight up, so θ = 90 degrees)
- g is the acceleration due to gravity, approximately 9.8 m/s^2
Plugging in the values:
h = (0^2 * sin^2 (90 degrees)) / (2 * 9.8 m/s^2)
h = 0 m
Therefore, the arrow does not rise to any height.