Robin exerts a force of 100 N in pulling back his bow by 0.5 m. He fires the arrow (mass=0.2 kg) vertically upwards.
a-How much energy is stored in his bow.
b-How high does the arrow go?
a) Stored energy = Work
= Force x (distance string is pulled)
b) M g H = stored energy
Solve for the height, H
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a- To calculate the amount of energy stored in Robin's bow, we can use the formula for potential energy. The potential energy stored in the bow is equal to the work done in pulling it back.
The work done is given by the formula:
Work = Force x Displacement x cos(theta)
Here, the force exerted by Robin is 100 N, and the displacement is 0.5 m. As the force is being applied in the same direction as the displacement, the angle (theta) between the force and displacement vectors is 0 degrees, and cos(0) = 1.
So, the work done is:
Work = 100 N x 0.5 m x 1 = 50 Joules
Therefore, the energy stored in the bow is 50 Joules.
b- To calculate the height reached by the arrow, we can use the principles of conservation of mechanical energy. Initially, the arrow has only potential energy stored in the bow. When it reaches its maximum height, all this potential energy is converted into kinetic energy.
The potential energy at maximum height is given by the formula:
Potential Energy = mass x acceleration due to gravity x height
Here, the mass of the arrow is 0.2 kg, and the acceleration due to gravity is approximately 9.8 m/s^2.
We can equate the potential energy at maximum height to the initial energy stored in the bow:
Potential Energy = 50 Joules
50 Joules = 0.2 kg x 9.8 m/s^2 x height
Solving for height:
height = 50 Joules / (0.2 kg x 9.8 m/s^2)
height ≈ 25.5 meters
Therefore, the arrow reaches a maximum height of approximately 25.5 meters.
a- To find the amount of energy stored in Robin's bow, we can use the formula for elastic potential energy:
Elastic 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 force exerted by Robin is 100 N, and the displacement of the bowstring is 0.5 m. The spring constant can be calculated using Hooke's Law:
F = k × x
Rearranging the equation, we find:
k = F / x
Substituting the values:
k = 100 N / 0.5 m = 200 N/m
Now we can calculate the energy stored in the bow:
Elastic Potential Energy = (1/2) × 200 N/m × (0.5 m)^2
= 50 J
Therefore, there is 50 joules of energy stored in Robin's bow.
b- To find the height the arrow reaches, we can use the principles of conservation of mechanical energy. At the highest point of its trajectory, the arrow will have no kinetic energy and all its initial potential energy will be converted to gravitational potential energy.
The initial potential energy of the arrow is given by:
Initial Potential Energy = m × g × h
where m is the mass of the arrow, g is the acceleration due to gravity (approximately 9.8 m/s^2), and h is the height the arrow reaches.
From part a, we know that the energy stored in the bow is 50 J. Therefore:
Initial Potential Energy = 50 J
Substituting the values:
50 J = 0.2 kg × 9.8 m/s^2 × h
Simplifying:
h = 50 J / (0.2 kg × 9.8 m/s^2)
= 25 m
Therefore, the arrow reaches a height of 25 meters.