Imagine a horizontal spring with an elastic constant of 26.5 N/m, held in place just above an air table. If a 2.00 kg mass traveling at 1.50 m/s slams axially into the end of the spring, by how much will it be compressed in bringing the mass to rest?
To find out how much the spring will be compressed, we can use the principle of conservation of energy. The initial kinetic energy of the mass will be transferred into the potential energy stored in the spring.
1. First, let's calculate the initial kinetic energy of the mass using the formula:
Kinetic Energy = (1/2) * mass * velocity^2
Given:
Mass (m) = 2.00 kg
Velocity (v) = 1.50 m/s
Plugging the values into the formula:
Kinetic Energy = (1/2) * 2.00 kg * (1.50 m/s)^2
2. The potential energy stored in the spring is given by the formula:
Potential Energy = (1/2) * k * x^2
Where:
k is the elastic constant of the spring, given as 26.5 N/m.
x is the compression of the spring, which we need to find.
3. Since energy is conserved, the initial kinetic energy is equal to the potential energy of the spring:
Kinetic Energy = Potential Energy
Setting the equations equal to each other:
(1/2) * 2.00 kg * (1.50 m/s)^2 = (1/2) * 26.5 N/m * x^2
4. Let's solve for x, the compression of the spring:
x^2 = [2 * (1.50 m/s)^2] / 26.5 N/m
Simplifying the equation:
x^2 = 0.169 m^2
5. Taking the square root of both sides to solve for x:
x ≈ √(0.169 m^2)
x ≈ 0.411 m
Therefore, the spring will be compressed by approximately 0.411 meters when the mass comes to rest.