A 280 g wood block is firmly attached to a very light horizontal spring, as shown in the figure . The block can slide along a table where the coefficient of friction is 0.30. A force of 22 N compresses the spring 18 cm. If the spring is released from this position, how far beyond its equilibrium position will it stretch at its first maximum extension?

Figure the stored energy at the beginning (compressed 18 cm).

Figure the friction energy it loses as it slides x+18 cm (mg*mu*(x+.18)

final stored energy= 1/2 k x^2 solve for x.

base equation: final stored energy= initial stored energy + friction energy lost.

30

it must be = 0.17

To determine how far beyond its equilibrium position the spring will stretch at its first maximum extension, we can use the principle of conservation of mechanical energy.

1. First, let's determine the potential energy stored in the spring when it is compressed. The potential energy stored in a spring is given by the equation:

Potential energy (PE) = (1/2) * k * x^2

where k is the spring constant and x is the displacement from the equilibrium position. Given that the spring is compressed by 18 cm (or 0.18 m) and the force applied is 22 N, we can rearrange Hooke's Law (F = kx) to find the spring constant (k):

k = F / x = 22 N / 0.18 m

2. With the spring constant known, we can calculate the potential energy stored in the spring when it is compressed:

PE = (1/2) * k * x^2 = (1/2) * (22 N / 0.18 m) * (0.18 m)^2

3. Now, let's consider the situation when the spring is released. At its first maximum extension, the potential energy is converted to kinetic energy (at maximum extension, the block momentarily comes to rest before moving back). So we have:

PE = KE

where KE is the kinetic energy of the block at maximum extension.

The kinetic energy of an object is given by the equation:

KE = (1/2) * m * v^2

where m is the mass of the block and v is its velocity.

4. The velocity at maximum extension can be calculated using conservation of mechanical energy. As the block comes to rest at maximum extension, its kinetic energy is zero. Therefore, the conservation of mechanical energy equation is:

PE_initial = PE_final

Initial potential energy = Final potential energy + final kinetic energy

Rearranging this equation, we get:

v^2 = 2 * (PE_initial - PE_final) / m

where PE_initial is the potential energy when the spring is compressed and PE_final is the potential energy when the block is at maximum extension.

5. Substituting the expressions for potential energy, mass, and solving for v^2:

v^2 = 2 * [(1/2) * (22 N / 0.18 m) * (0.18 m)^2 - 0] / 0.280 kg

6. Now that we have the velocity at maximum extension (v), we can find the displacement from the equilibrium position (maximum extension):

v = dx / dt

where dx is the displacement from the equilibrium position and dt is the time taken.

Since we are considering instantaneous maximum extension, the time taken (dt) is very small.

Therefore, dx ≈ v * dt

7. Finally, we need to convert the displacement (dx) from meters to centimeters (cm) since the given value is in centimeters. Multiply dx by 100 to obtain the answer in centimeters.

By following these steps, you can calculate how far beyond its equilibrium position the spring will stretch at its first maximum extension.