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 out how much the spring is compressed by the brick, we can use the principle of conservation of mechanical energy. This principle states that the total mechanical energy of a system remains constant if no external forces are acting on it.

The total mechanical energy of the system can be calculated as the sum of the potential energy (PE) and the kinetic energy (KE) at any point in time. Initially, the brick has only potential energy, which can be calculated using the formula PE = mgh, 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 initial height (6 m). Therefore, the initial potential energy of the brick is PE = 2 kg * 9.8 m/s^2 * 6 m = 117.6 J.

When the brick reaches the spring, all of its potential energy is converted into elastic potential energy stored in the compressed spring. The formula for elastic potential energy is PE = (1/2)kx^2, where k is the spring constant (47 N/m) and x is the compression distance of the spring.

Since the brick comes to a stop, all of its kinetic energy is converted into elastic potential energy. Therefore, we can equate the initial potential energy to the final elastic potential energy as follows:

117.6 J = (1/2) * 47 N/m * x^2

Simplifying the equation, we have:

235.2 J = 47 N/m * x^2

Now, we can solve for x:

x^2 = 235.2 J / 47 N/m

x^2 = 5 m^2

Taking the square root of both sides, we get:

x = ±√5 m

Since the brick compresses the spring, the negative root is not physically meaningful in this context. Therefore, the spring is compressed by √5 m, which is approximately 2.24 m.

Hence, the spring is compressed by approximately 2.24 meters.