A cork gun contains a spring whose spring constant is 11.0 N/m. The spring is compressed by a distance DX = 9.8 cm and used to propel a cork of mass 6.60 g from the gun. Assuming the cork is released when the spring passes through its equilibrium position (Xeq), what is the speed of the cork as it is released from the spring?
Suppose now that the cork temporarily sticks to the spring, causing the spring to extend 4.5 cm beyond its equilibrium position before separation occurs. What is the speed of the cork as it is released from the spring in this case?
To find the speed of the cork as it is released from the spring in both cases, we can use the principle of conservation of mechanical energy. The total mechanical energy of the system remains constant throughout the motion.
1. In the first case, where the cork is released when the spring passes through its equilibrium position (Xeq):
The total mechanical energy of the system is given by the sum of the potential energy stored in the spring and the kinetic energy of the cork.
Potential energy of the spring, Us = (1/2)kX^2
where k is the spring constant and X is the displacement from the equilibrium position.
Us = (1/2) * 11.0 N/m * (0.098 m)^2
Kinetic energy of the cork, Kc = (1/2)mv^2
where m is the mass of the cork and v is its speed.
Since the total mechanical energy is conserved, we can equate the potential energy to the kinetic energy and solve for v:
(1/2)kX^2 = (1/2)mv^2
Plugging in the values:
(1/2) * 11.0 N/m * (0.098 m)^2 = (1/2) * 0.00660 kg * v^2
Simplifying and solving for v:
v = sqrt((11.0 N/m * (0.098 m)^2) / 0.00660 kg)
2. In the second case, where the cork sticks to the spring and it extends 4.5 cm beyond its equilibrium position before separation:
In this case, when the cork temporarily sticks to the spring, both the cork and the spring move together until separation occurs. The cork-spring system behaves like a mass-spring system.
The potential energy of the spring is given by (1/2)kX^2, where X is the displacement of the spring from its equilibrium position.
In this case, X = Xeq + 0.045 m.
Setting the potential energy of the spring equal to the kinetic energy of the cork, we have:
(1/2)k(Xeq + 0.045 m)^2 = (1/2)mv^2
Plugging in the values:
(1/2) * 11.0 N/m * (0.098 m + 0.045 m)^2 = (1/2) * 0.00660 kg * v^2
Simplifying and solving for v:
v = sqrt((11.0 N/m * (0.098 m + 0.045 m)^2) / 0.00660 kg)
By substituting the given values into the respective equations, you can calculate the speed of the cork in each case.
To find the speed of the cork as it is released from the spring in the first case, we can use the principle of conservation of mechanical energy:
1. Calculate the potential energy stored in the compressed spring:
PE = (1/2)k(DX)^2
where k is the spring constant and DX is the compression distance.
PE = (1/2)(11.0 N/m)(0.098 m)^2
PE = 0.053964 J
2. Calculate the kinetic energy of the cork as it is released:
KE = PE
So, KE = 0.053964 J
3. Calculate the velocity of the cork:
KE = (1/2)mv^2
where m is the mass of the cork and v is its velocity.
Rearranging the equation: v = sqrt((2KE)/m)
v = sqrt((2(0.053964 J))/(6.60 g))
v = sqrt((0.107928 J)/(0.00660 kg))
v ≈ 4.16 m/s
So, the speed of the cork as it is released from the spring in the first case is approximately 4.16 m/s.
In the second case, where the cork temporarily sticks to the spring causing an additional displacement, we can use the conservation of mechanical energy again to find the speed of the cork:
1. Calculate the potential energy stored in the stretched spring:
PE = (1/2)k(DX+DS)^2
where DS is the additional displacement beyond the equilibrium position.
PE = (1/2)(11.0 N/m)(0.098 m + 0.045 m)^2
PE = 0.071478 J
2. Calculate the kinetic energy of the cork as it is released:
KE = PE
So, KE = 0.071478 J
3. Calculate the velocity of the cork:
v = sqrt((2KE)/m)
v = sqrt((2(0.071478 J))/(6.60 g))
v = sqrt((0.142956 J)/(0.00660 kg))
v ≈ 4.81 m/s
So, the speed of the cork as it is released from the spring in the second case is approximately 4.81 m/s.