An archer puts a 0.26-kg arrow to the bow-string. An average force of 200 N is exerted to draw the string back 1.4 m.
(a) Assuming no frictional loss, with what speed does the arrow leave the bow?
m/s
(b) If the arrow is shot straight up, how high does it rise?
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.
To find the answers to these questions, we need to use the principles of work and energy conservation. We'll start by calculating the potential energy stored in the bowstring as it is pulled back.
(a) Finding the speed of the arrow when it leaves the bow:
1. Calculate the potential energy stored in the bowstring:
The potential energy is given by the formula: potential energy = force × distance
In this case, the force exerted is 200 N and the distance is 1.4 m.
So, potential energy = 200 N × 1.4 m
2. Use the principle of energy conservation:
Since there is no frictional loss, the potential energy stored in the bowstring is converted into kinetic energy when the arrow is released.
The kinetic energy is given by the formula: kinetic energy = (1/2) × mass × speed^2
The mass of the arrow is given as 0.26 kg.
3. Equate the potential energy to the kinetic energy and solve for the speed:
(1/2) × mass × speed^2 = potential energy
Substitute the values: (1/2) × 0.26 kg × speed^2 = (200 N × 1.4 m)
Solve the equation for speed.
(b) Finding the maximum height reached by the arrow when shot straight up:
1. Use the principle of conservation of energy:
At its maximum height, all the kinetic energy of the arrow is converted into potential energy.
We can equate the kinetic energy at the start with the potential energy at the highest point.
2. Calculate the potential energy at the maximum height:
The potential energy at the highest point is given by the formula: potential energy = mass × gravitational acceleration × height
The gravitational acceleration is approximately 9.8 m/s^2.
3. Equate the potential energy to the initial kinetic energy and solve for the height:
(1/2) × mass × speed^2 = mass × gravitational acceleration × height
Substitute the values and solve the equation for height.
By following these steps, we can calculate the speed of the arrow leaving the bow and the maximum height it reaches when shot straight up.