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?
m
1.
W =F•x =200•1.4 =280 J,
W =KE = m•v²/2,
v =sqrt(2•W/m) = sqrt(2•280/0.26) =46.4 m/s.
KE =PE = m•g•h
h = KE/m•g = 280/0.26•9.8 ≈110 m.
2.
Sometimes I’ve found the answers of this problem based on the following solution
PE =kx²/2 = KE =mv²/2.
This solution gives v =sqrt(k•x²/m) = sqrt(F•x²/x•m) = sqrt(F•x/m) = 32.8 m/s.
And the height from m•g•h =m•v²/2 is
h =v²/2•g = 54.95 m.
But I believe that this solution is incorrect because we have given the magnitude of the average (!) force
So the first solution is quite the thing.
To answer part (a) of the question, we can use the principle of conservation of mechanical energy. The energy stored in the bow is transferred to the arrow as it is released. We can find the speed of the arrow using the following steps:
Step 1: Calculate the potential energy stored in the bow when the string is drawn back.
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 (approximately 9.8 m/s²), and h is the height from which the arrow is released.
Given:
Mass of the arrow (m) = 0.26 kg
Height from which the arrow is released (h) = 0.00 m (assuming it is released horizontally)
PE = mgh = (0.26 kg)(9.8 m/s²)(0.00 m) = 0 J (since h = 0 m)
Step 2: Calculate the work done by the bowstring in pulling the arrow back.
The work done (W) is given by the equation W = Fd, where F is the force applied and d is the distance over which the force is applied.
Given:
Force applied (F) = 200 N
Distance over which the force is applied (d) = 1.4 m
W = Fd = (200 N)(1.4 m) = 280 J
Step 3: Calculate the kinetic energy of the arrow as it leaves the bow.
The kinetic energy (KE) is given by the equation KE = 0.5mv², where m is the mass of the arrow and v is its velocity.
Since energy is conserved, the work done by the bowstring (W) is equal to the kinetic energy of the arrow (KE).
280 J = 0.5(0.26 kg)v²
Simplifying the equation, we find:
v² = (2 × 280 J) / (0.26 kg) = 2153.846 m²/s²
Taking the square root of both sides, we find:
v ≈ √(2153.846 m²/s²) ≈ 46.42 m/s
Therefore, the speed at which the arrow leaves the bow is approximately 46.42 m/s.
To answer part (b) of the question, we can use the kinematic equations to find the maximum height reached by the arrow when shot straight up. The equation we can use is:
vf² = vi² + 2ad
Where:
- vf is the final velocity (0 m/s at the maximum height),
- vi is the initial velocity (46.42 m/s, as found in part (a)),
- a is the acceleration due to gravity (-9.8 m/s², since the arrow is moving in the opposite direction),
- d is the displacement or change in height from the starting point.
We can rearrange the equation to solve for d:
d = (vf² - vi²) / (2a) = (0 - (46.42 m/s)²) / (2 × -9.8 m/s²) = -47.84 m
Since the height cannot be negative, we take the absolute value:
The arrow rises to a height of approximately 47.84 m before falling back down.
Therefore, the height to which the arrow rises is approximately 47.84 m.