A 2 kg brick falls at 5.1 m/s from a height of 6 m above a spring that has a spring constant of 47 N/m. The brick strikes and compresses the spring until it stops moving. Ignore friction effects.
How much is the spring compressed by the brick?
Assume that there is no Eg after the brick comes into contact with the spring.
To find the amount by which the spring is compressed by the brick, we can use the principles of conservation of energy and Hooke's Law.
First, let's calculate the potential energy (PE) of the brick when it is at a height of 6 m above the spring. The potential energy is given by the equation: PE = m * g * h, where m is the mass of the brick (2 kg), g is the acceleration due to gravity (9.8 m/s^2), and h is the height (6 m).
PE = 2 kg * 9.8 m/s^2 * 6 m = 117.6 kg m^2/s^2
Next, let's calculate the initial kinetic energy (KE) of the brick when it is about to strike the spring. The initial kinetic energy is given by the equation: KE = 0.5 * m * v^2, where v is the velocity of the brick (5.1 m/s).
KE = 0.5 * 2 kg * (5.1 m/s)^2 = 26.01 kg m^2/s^2
Since the problem states that there is no Eg (gravitational potential energy) after the brick comes into contact with the spring, all the initial potential energy is converted into the elastic potential energy stored in the compressed spring.
Elastic potential energy (PE_spring) is given by the equation: PE_spring = 0.5 * k * x^2, where k is the spring constant (47 N/m) and x is the compression distance of the spring.
We can set up an equation equating the initial potential energy to the elastic potential energy:
PE = PE_spring
117.6 kg m^2/s^2 = 0.5 * 47 N/m * x^2
Now we can solve for the compression distance (x):
2 * 9.8 * 6 = 47 * x^2
117.6 = 47 * x^2
x^2 = 117.6 / 47
x^2 ≈ 2.496
x ≈ √2.496
x ≈ 1.58 m
Therefore, the spring is compressed by approximately 1.58 m when the brick stops moving.
Note: The negative sign in the answer is ignored since we are only interested in the magnitude of the compression.