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!
To find the speed of the cork as it is released from the spring, we can use the principle of conservation of mechanical energy.
1. In the first case, where the cork is released when the spring passes through its equilibrium position (Xeq):
The potential energy stored in the spring when it is compressed by a distance ∆X is given by:
PE = (1/2)k(∆X)^2
where k is the spring constant.
Given that the spring constant (k) is 18.0 N/m and the distance (∆X) is 6.0 cm (or 0.06 m), we can calculate the potential energy:
PE = (1/2)(18.0 N/m)(0.06 m)^2
= 0.0324 J
Since the potential energy is converted into kinetic energy when the cork is released, we can equate the two:
KE = PE
The kinetic energy is given by:
KE = (1/2)mv^2
where m is the mass of the cork and v is its speed.
Given that the mass (m) of the cork is 7.66 g (or 0.00766 kg), we can rearrange the equation to solve for the speed:
v^2 = (2KE)/m
v^2 = (2PE)/m
v = √[(2PE)/m]
v = √[(2 * 0.0324 J) / 0.00766 kg]
v ≈ 6.31 m/s
Therefore, the speed of the cork as it is released from the spring in the first case is approximately 6.31 m/s.
2. In the second case, where the cork temporarily sticks to the spring and causes it to extend 3.0 cm beyond its equilibrium position before separation occurs:
The potential energy stored in the extended spring when the cork sticks to it is given by:
PE = (1/2)k(∆X)^2
where ∆X is the extension of the spring beyond its equilibrium position.
Given that the spring constant (k) is still 18.0 N/m and the extension (∆X) is 3.0 cm (or 0.03 m), we can calculate the potential energy:
PE = (1/2)(18.0 N/m)(0.03 m)^2
= 0.0081 J
The kinetic energy of the cork when it separates from the spring is given by the same equation as before:
KE = (1/2)mv^2
Using the same mass (m) of the cork as before, we can solve for the speed:
v^2 = (2KE)/m
v^2 = (2PE)/m
v = √[(2PE)/m]
v = √[(2 * 0.0081 J) / 0.00766 kg]
v ≈ 2.09 m/s
Therefore, the speed of the cork as it is released from the spring in the second case is approximately 2.09 m/s.
To find the speed of the cork as it is released from the spring, we can use the principle of conservation of mechanical energy. The initial potential energy stored in the compressed spring is converted into the kinetic energy of the cork.
First, we need to convert the mass of the cork from grams to kilograms. The mass of the cork is given as 7.66 g, which is equal to 0.00766 kg.
Now, let's calculate the potential energy stored in the spring when it is compressed. The potential energy stored in a spring is given by the formula:
PE = (1/2)k∆X^2
where PE is the potential energy, k is the spring constant, and ∆X is the distance the spring is compressed.
Substituting the given values, we have:
PE = (1/2) * 18.0 N/m * (0.06 m)^2 = 0.0324 J
Since the cork is initially at rest, all of the potential energy stored in the spring is converted into kinetic energy when it is released. Therefore, the kinetic energy of the cork can be calculated as:
KE = PE
Now, we can use the formula for kinetic energy to find the speed of the cork:
KE = (1/2)mv^2
Rearranging the formula, we have:
v^2 = 2KE / m
Substituting the calculated values, we get:
v^2 = 2 * 0.0324 J / 0.00766 kg = 84.10 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 9.18 m/s
Therefore, the speed of the cork as it is released from the spring is approximately 9.18 m/s.
Now, let's consider the case where the cork temporarily sticks to the spring, causing it to extend 3.0 cm beyond its equilibrium position before separation occurs.
In this case, we need to find the new potential energy stored in the extended spring. The distance the spring is extended is 3.0 cm, which is equal to 0.03 m.
Using the formula for potential energy, we have:
PE = (1/2)k∆X^2
Substituting the given values, we get:
PE = (1/2) * 18.0 N/m * (0.03 m)^2 = 0.009 J
Since the cork is initially at rest, all of the potential energy stored in the extended spring is converted into kinetic energy when it is released. Therefore, the kinetic energy of the cork can be calculated as:
KE = PE
Using the formula for kinetic energy and substituting the values, we have:
KE = (1/2)mv^2
v^2 = 2KE / m
v^2 = 2 * 0.009 J / 0.00766 kg = 2.34 m^2/s^2
Taking the square root of both sides, we find:
v ≈ 1.53 m/s
Therefore, the speed of the cork as it is released from the spring when it temporarily sticks and extends beyond its equilibrium position is approximately 1.53 m/s.
F = (0.06m/1m) * 18N = 1.08 N.
a = F/m = 1.08/7.66*10^-3 = 141 m/s^2.
V^2 = Vo^2 + 2a>d
V^2 = 0 + 282*0.06 = 16.92
V = 4.11 m/s.
V^2 = 4.11^2 + 282*0.03 = 25.35
V = 5.03 m/s.