The launching mechanism of a toy gun consists of a spring of unknown spring constant, as shown in the figure below. If the spring is compressed a distance of 0.160 m and the gun fired vertically as shown, the gun can launch a 21.0-g projectile from rest to a maximum height of 26.0 m above the starting point of the projectile.
a)Neglecting all resistive frces, determine the spring constant.
b)Neglecting all resistive forces, find the speed of the projectile as it moves through the equilibrium position of the spring (where x = 0), as shown in Figure (b).
To solve this problem, we will use the principles of conservation of mechanical energy and Hooke's Law.
a) To determine the spring constant, we need to find the potential energy stored in the spring when it is compressed.
The potential energy stored in a spring is given by the equation:
PEspring = (1/2)kx^2
Where PEspring is the potential energy stored in the spring, k is the spring constant, and x is the compression distance.
Given:
Compression distance, x = 0.160 m
Potential energy, PEspring = 21.0 g * 9.8 m/s^2 * 26.0 m (converted to joules) = 52.956 J
Substituting these values into the equation, we have:
52.956 J = (1/2)k * (0.160 m)^2
Simplifying the equation:
104.82 = k * 0.0256
Solving for k:
k = 104.82 J / 0.0256 m^2
k ≈ 4096.09 N/m
Therefore, the spring constant is approximately 4096.09 N/m.
b) To find the speed of the projectile at the equilibrium position (where x = 0), we can equate the potential energy of the spring to the kinetic energy of the projectile when it reaches the maximum height.
The potential energy at the maximum height is given by:
PEgravity = m * g * h
Where m is the mass of the projectile, g is the acceleration due to gravity, and h is the maximum height.
Given:
Mass of the projectile, m = 21.0 g = 0.021 kg
Acceleration due to gravity, g = 9.8 m/s^2
Maximum height, h = 26.0 m
Substituting these values into the equation, we have:
PEgravity = 0.021 kg * 9.8 m/s^2 * 26.0 m
PEgravity ≈ 53.692 J
The potential energy at the maximum height is equal to the initial potential energy of the spring when the projectile is at the equilibrium position. Therefore, we equate this to the kinetic energy of the projectile when it reaches the equilibrium position:
KE = (1/2)mv^2
Where KE is the kinetic energy of the projectile, m is the mass of the projectile, and v is the velocity of the projectile.
Substituting the values into the equation, we have:
53.692 J = (1/2) * 0.021 kg * v^2
Simplifying the equation:
107.384 = 0.021 v^2
Solving for v:
v^2 = 107.384 / 0.021
v^2 ≈ 5118.1
Taking the square root of both sides:
v ≈ √5118.1
v ≈ 71.57 m/s
Therefore, the speed of the projectile as it moves through the equilibrium position is approximately 71.57 m/s.