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?

To find how far beyond its equilibrium position the spring will stretch at its first maximum extension, we can use the concept of potential energy.

First, let's determine the spring constant, k. The spring constant is a measure of how stiff the spring is and can be calculated using Hooke's Law:

F = kx

Where F is the force applied to the spring and x is the displacement of the spring from its equilibrium position.

In this case, the force applied to the spring is 22 N, and the displacement is 18 cm (or 0.18 m). Thus, we can rearrange the equation to solve for k:

k = F / x

k = 22 N / 0.18 m
k ≈ 122.22 N/m

Now that we have the spring constant, we can determine 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)kx²

Where PE is the potential energy and x is the displacement of the spring from its equilibrium position.

In this case, the displacement is 0.18 m, so we can calculate the potential energy:

PE = (1/2) * 122.22 N/m * (0.18 m)²
PE ≈ 3.31 J

Since energy is conserved, this potential energy will be converted into kinetic energy when the spring is released. At the maximum extension, all the potential energy will be converted into kinetic energy.

At maximum extension, the kinetic energy is given by:

KE = (1/2)mv²

Where KE is the kinetic energy, m is the mass of the block, and v is the velocity of the block at maximum extension.

In this case, the mass of the block is 280 g (or 0.28 kg).

Now, we can equate the potential energy to the kinetic energy:

PE = KE

(1/2)mv² = (1/2)kx²

Canceling the common factors:

mv² = kx²

Plugging in the values:

(0.28 kg)v² = (122.22 N/m)(0.18 m)²

Rearranging and solving for v:

v² = (122.22 N/m)(0.18 m)² / 0.28 kg
v ≈ 1.77 m/s

Now, we know the velocity of the block at maximum extension. To find the distance it will stretch beyond the equilibrium position, we can use the concept of kinetic friction.

The force of kinetic friction can be calculated using the formula:

fk = μk * N

Where fk is the force of kinetic friction, μk is the coefficient of friction (0.30), and N is the normal force.

In this case, since the block is horizontal and there is no vertical acceleration, the normal force is equal to the weight of the block:

N = mg

N = (0.28 kg)(9.8 m/s²)
N ≈ 2.74 N

Now, we can calculate the force of kinetic friction:

fk = (0.30)(2.74 N)
fk ≈ 0.822 N

The force of kinetic friction will act in the opposite direction of the motion. Therefore, we can use the work-energy principle to find the distance the block will stretch beyond the equilibrium position:

fk * d = (1/2)mv²

(0.822 N)d = (1/2)(0.28 kg)(1.77 m/s)²

Simplifying and solving for d:

d = [(1/2)(0.28 kg)(1.77 m/s)²] / (0.822 N)
d ≈ 0.94 m

Therefore, the spring will stretch approximately 0.94 meters (or 94 cm) beyond its equilibrium position at its first maximum extension.