An archer locks a 0.20kg arrow on a bowstring. Then the archer exerts an average force of 110N to draw the string back 0.60 m . Assume that friction is negligible .
What speed does the bow give to the arrow?
If the arrow is shot vertically upward,how high will it rise?
W =KE
F•s =mv(o)^2/2
v(o) = sqrt(2•F•s/m) = sqrt (2•110•0.6/0.2)= 25.7 m/s
v = v(o) - g•t,
v=0 => t = v(o)/g =25.7/9.8 =2.6 s
h =v(o) •t –g•t^2/2 = 25.7•2.6 - 9.8•(2.6)^2/2 = 33.1 m
Well, that's quite an archery challenge! Let's get to it!
To find the speed the bow gives to the arrow, we can use the principle of conservation of energy. The energy stored in the bow when it's drawn back is transferred to the arrow. So, we can equate the potential energy stored in the bow to the kinetic energy of the arrow.
The potential energy stored in the bow is given by the equation:
Potential Energy = (1/2) * k * x^2
Where k is the spring constant of the bow and x is the displacement (0.60 m). Since the force exerted by the archer is 110N and the displacement is 0.60 m, we can find the spring constant using Hooke's Law:
F = k * x
110N = k * 0.60m
k = 110N / 0.60m
k ≈ 183.33 N/m
Now that we have the spring constant, we can calculate the potential energy stored in the bow:
Potential Energy = (1/2) * 183.33 N/m * (0.60 m)^2
Potential Energy ≈ 39.99 Joules
Since energy is conserved, this potential energy will be converted into kinetic energy when the arrow is released. The kinetic energy of the arrow is given by:
Kinetic Energy = (1/2) * m * v^2
Where m is the mass of the arrow and v is its velocity. Rearranging the equation, we can solve for velocity:
v = √(2 * kinetic energy / mass)
Plugging in the values:
v = √(2 * 39.99 J / 0.20 kg)
v = √(399.8 (kg * m^2 / s^2) / 0.20 kg)
v = √(1999 m^2 / s^2)
v ≈ 44.72 m/s
So, the bow gives the arrow a speed of approximately 44.72 m/s! Now that's some impressive speed for an arrow!
Moving on to the second part of the question, if the arrow is shot vertically upwards, it will reach a maximum height when its velocity becomes zero. At this point, all of its initial kinetic energy is converted into potential energy.
Using the principle of conservation of energy again, we can equate the initial kinetic energy of the arrow to the potential energy at its maximum height:
(1/2) * m * v^2 = m * g * h
Where m is the mass of the arrow, v is its velocity (44.72 m/s), g is the acceleration due to gravity (9.8 m/s^2), and h is the maximum height.
Simplifying the equation:
(1/2) * (0.20 kg) * (44.72 m/s)^2 = (0.20 kg) * (9.8 m/s^2) * h
Solving for h:
h = [(0.20 kg) * (44.72 m/s)^2] / [(0.20 kg) * (9.8 m/s^2)]
h = (0.20 kg * 2000.4384 m^2/s^2) / (0.2 kg * 9.8 m/s^2)
h ≈ 204.09 m
So, if the arrow is shot vertically upward, it will rise to approximately 204.09 meters. That's pretty high! Just make sure not to stand underneath it when it comes back down. Safety first, after all!
To find the speed that the bow gives to the arrow, we can use the principle of conservation of mechanical energy. The work done by the archer in drawing the bowstring back will be equal to the potential energy stored in the bowstring, which will be converted to the kinetic energy of the arrow once released.
Step 1: Calculate the potential energy stored in the bowstring.
The potential energy stored in the bowstring is given by the formula:
Potential Energy = (1/2) k x^2,
where k is the spring constant and x is the displacement of the bowstring.
In this case, the potential energy stored in the bowstring is:
Potential Energy = (1/2) (110 N/m) (0.60 m)^2
Potential Energy = 11.22 Joules
Step 2: Calculate the kinetic energy of the arrow.
The kinetic energy of the arrow can be calculated using the formula:
Kinetic Energy = (1/2) m v^2,
where m is the mass of the arrow and v is the velocity of the arrow.
In this case, the mass of the arrow is 0.20 kg.
Step 3: Equate the potential energy to the kinetic energy.
Since the potential energy is converted to kinetic energy, we have:
Potential Energy = Kinetic Energy
11.22 J = (1/2) (0.20 kg) v^2
Step 4: Solve for the velocity of the arrow.
Rearrange the equation to solve for v:
v^2 = (2 * 11.22 J) / 0.20 kg
v^2 = 112.2 m^2/s^2
v ≈ 10.6 m/s
Therefore, the bow gives the arrow a speed of approximately 10.6 m/s.
To find the maximum height the arrow will reach when shot vertically upward, we can use the equation for projectile motion.
Step 1: Calculate the time of flight.
The time of flight can be calculated using the equation:
Time = (2 * vertical component of velocity) / acceleration due to gravity
The vertical component of velocity is the initial velocity of the arrow, which is 10.6 m/s.
Step 2: Calculate the maximum height.
The maximum height can be calculated using the equation:
Height = (vertical component of velocity)^2 / (2 * acceleration due to gravity)
The acceleration due to gravity is approximately 9.8 m/s^2.
Step 3: Substitute the values into the equations.
Using the formulas mentioned above, we can calculate the time of flight and maximum height:
Time = (2 * 10.6 m/s) / 9.8 m/s^2 ≈ 2.17 s
Height = (10.6 m/s)^2 / (2 * 9.8 m/s^2) ≈ 5.52 m
Therefore, the arrow will rise to a maximum height of approximately 5.52 meters when shot vertically upward.
To find the speed that the bow gives to the arrow, we can use the principle of work and energy. The work done by the archer in drawing the bowstring back is equal to the potential energy stored in the bow. This potential energy is then converted into kinetic energy when the arrow is released.
1. To calculate the potential energy stored in the bow, we use the formula for potential energy:
Potential Energy = (1/2) * k * x^2
Where:
k = spring constant of the bow
x = displacement of the bowstring (0.60 m in this case)
Since the spring constant is not given, we can calculate it using Hooke's Law:
F = -k * x
Rearranging the equation, we find:
k = -F / x
Plugging in the values, we get:
k = -110 N / 0.60 m
2. Now that we have the spring constant, we can calculate the potential energy:
Potential Energy = (1/2) * k * x^2
= (1/2) * (-110 N / 0.60 m) * (0.60 m)^2
3. The potential energy is equal to the kinetic energy of the arrow when it is released:
Kinetic Energy = Potential Energy
Since kinetic energy is given by the equation:
Kinetic Energy = (1/2) * m * v^2
Where:
m = mass of the arrow (0.20 kg)
v = velocity of the arrow
We can rearrange the equation to solve for v:
v = sqrt((2 * Potential Energy) / m)
Plugging in the values, we get:
v = sqrt((2 * Potential Energy) / 0.20 kg)
Now, for the second question - If the arrow is shot vertically upward, how high will it rise?
We can use the conservation of mechanical energy to find the maximum height reached by the arrow. At its highest point, all of the kinetic energy of the arrow is converted into potential energy.
4. We can calculate the potential energy at the maximum height using the equation:
Potential Energy = m * g * h
Where:
m = mass of the arrow (0.20 kg)
g = acceleration due to gravity (9.8 m/s^2)
h = maximum height reached
5. Since the potential energy at the maximum height is equal to the kinetic energy of the arrow when it was shot upward:
Potential Energy = Kinetic Energy
We can set up the equation:
(1/2) * m * v^2 = m * g * h
Rearranging the equation, we find:
h = (1/2) * v^2 / g
Plugging in the values we calculated for v and g, we get:
h = (1/2) * (sqrt((2 * Potential Energy) / 0.20 kg))^2 / 9.8 m/s^2
Simply calculate the values in steps 3 and 5 to get the final answers for the speed of the arrow and the maximum height it will reach.