A cork gun contains a spring whose spring constant is 18.0 N/m. The spring is compressed by a distance ∆X = 6.0 cm and used to propel a cork of mass 7.66 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 3.0 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?
Thanks!
a. PE in spring: 1/2 k x^2
KE in cork at equilibrium: 1/2 mv^2
set them equal, solve for v.
b. KE cork+PEspsring(3cm)=1/2 k (.06)^2
1/2 mv^2+1/2 k(.03)^2=1/2 k (.06)^2
check that.
To solve this problem, we can use the principle of conservation of energy. The initial potential energy stored in the spring is converted into the kinetic energy of the cork.
1. Calculate the initial potential energy stored in the spring:
The spring constant is given as k = 18.0 N/m, and the spring is compressed by a distance ∆X = 6.0 cm = 0.06 m. Using the formula for potential energy in a spring, the initial potential energy (U_i) is given by:
U_i = (1/2) * k * ∆X^2
= (1/2) * 18.0 N/m * (0.06 m)^2
2. Calculate the mass of the cork in kilograms:
The mass of the cork is given as 7.66 g = 0.00766 kg.
3. Calculate the speed of the cork as it is released from the spring:
Since the potential energy is converted into kinetic energy, we can equate U_i to the kinetic energy (K) of the cork.
U_i = K
(1/2) * k * ∆X^2 = (1/2) * m * v^2
Here, v represents the speed of the cork after it is released from the spring.
Let's go through the calculations for both cases:
Case 1: The cork is released when the spring passes through its equilibrium position.
1. U_i = (1/2) * 18.0 N/m * (0.06 m)^2
2. The mass of the cork, m = 0.00766 kg.
3. Equating potential energy to kinetic energy: (1/2) * 18.0 N/m * (0.06 m)^2 = (1/2) * 0.00766 kg * v^2
Solve for v: v^2 = [(1/2) * 18.0 N/m * (0.06 m)^2] / (0.00766 kg)
v = √{[(1/2) * 18.0 N/m * (0.06 m)^2] / (0.00766 kg)}
Now, let's move on to Case 2: The cork temporarily sticks to the spring, causing the spring to extend 3.0 cm beyond its equilibrium position before separation occurs.
1. U_i = (1/2) * 18.0 N/m * (0.06 m)^2
2. The mass of the cork, m = 0.00766 kg.
3. Equating potential energy to kinetic energy: (1/2) * 18.0 N/m * (0.06 m)^2 = (1/2) * 0.00766 kg * v^2
Solve for v: v^2 = [(1/2) * 18.0 N/m * (0.06 m + 0.03 m)^2] / (0.00766 kg)
v = √{[(1/2) * 18.0 N/m * (0.06 m + 0.03 m)^2] / (0.00766 kg)}
Calculating the expressions above will give you the specific values for the speed of the cork in each case.
To find the speed of the cork when it is released from the spring, we can use the principle of conservation of mechanical energy. In both cases, the initial mechanical energy stored in the compressed spring is converted into the kinetic energy of the cork as it is released.
1. Case 1: When the cork is released from the spring at its equilibrium position (∆X = 0):
The potential energy stored in the compressed spring, Us = (1/2)k∆X^2
Here, k is the spring constant and ∆X is the compression distance.
The potential energy stored in the spring is converted into the kinetic energy of the cork when released. The kinetic energy of the cork, Ek = (1/2)mv^2
Here, m is the mass of the cork, and v is its velocity.
Using the conservation of mechanical energy, we can equate the potential energy of the spring to the kinetic energy of the cork.
Us = Ek
(1/2)k∆X^2 = (1/2)mv^2
Now, let's substitute the given values:
k = 18.0 N/m (spring constant)
∆X = 6.0 cm = 0.06 m (compression distance)
m = 7.66 g = 0.00766 kg (mass of the cork)
Plugging in these values, we can solve for the speed v in equation (1).
(1/2)(18.0 N/m)(0.06 m)^2 = (1/2)(0.00766 kg)v^2
Simplifying the equation, we get:
0.0324 Nm = 0.0000295 kg v^2
Dividing both sides of the equation by 0.0000295 kg gives:
v^2 = 1.101 Ns^2/kg
Taking the square root of both sides, we get:
v = 1.05 m/s (approximately)
Therefore, the speed of the cork when released from the spring at its equilibrium position is 1.05 m/s.
2. Case 2: When the cork temporarily sticks to the spring and the spring extends 3.0 cm (∆X = -0.03 m) beyond its equilibrium position before separation occurs:
In this case, we need to consider the potential energy stored in the extended spring.
The potential energy of the extended spring, Us = (1/2)k∆X^2
Here, k is the spring constant, and ∆X is the extension distance.
The potential energy stored in the extended spring is converted into the kinetic energy of the cork when the separation occurs.
Using the conservation of mechanical energy, we can equate the potential energy of the extended spring to the kinetic energy of the cork.
Us = Ek
(1/2)k∆X^2 = (1/2)mv^2
Given values:
k = 18.0 N/m (spring constant)
∆X = -0.03 m (extension distance)
m = 7.66 g = 0.00766 kg (mass of the cork)
Substituting these values into equation (2), we can find the speed v.
(1/2)(18.0 N/m)(-0.03 m)^2 = (1/2)(0.00766 kg)v^2
Simplifying the equation, we get:
-0.01215 Nm = 0.0000295 kg v^2
Dividing both sides of the equation by 0.0000295 kg gives:
v^2 = -0.412 Ns^2/kg
Note that we have a negative value for the square root. Since speed cannot be negative, we can conclude that in this case, the cork does not have a valid release speed.