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?

.0134

nope...16.7 cm

To determine how far beyond its equilibrium position the spring will stretch at its first maximum extension, we need to consider the forces acting on the wood block.

First, let's calculate the spring constant (k) of the spring using Hooke's law:

F = kx

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

Given:
Force (F) = 22 N
Displacement (x) = 18 cm = 0.18 m

So, k = F / x = 22 N / 0.18 m = 122.22 N/m

Now, let's move on to the forces acting on the wood block when it is released from its compressed position.

The restoring force of the spring (Fr) will oppose the displacement of the block. This force can be calculated using the equation:

Fr = -kx

Where x is the displacement of the block from its equilibrium position.

The maximum extension occurs when the block comes to a stop due to the force of kinetic friction opposing its motion. The frictional force (Ff) can be calculated using:

Ff = μN

Where μ is the coefficient of friction and N is the normal force.

The normal force (N) can be calculated as the weight of the block (mg) where g is the acceleration due to gravity (approximately 9.8 m/s^2).

Given:
Mass (m) = 280 g = 0.28 kg
Coefficient of friction (μ) = 0.30

N = mg = 0.28 kg × 9.8 m/s^2 = 2.744 N

Now, let's calculate the frictional force:

Ff = μN = 0.30 × 2.744 N = 0.823 N

The net force acting on the block is the sum of the spring force and the frictional force:

Net force (Fnet) = Fr + Ff

At the maximum extension, the net force is zero since the block is at equilibrium:

Fnet = 0

Setting the net force equation equal to zero:

0 = -kx + Ff

Rearranging the equation:

kx = Ff

Substituting the known values:

122.22 N/m × x = 0.823 N

Solving for x:

x = 0.823 N / 122.22 N/m = 0.0067 m = 6.7 mm

Therefore, the spring will stretch approximately 6.7 mm beyond its equilibrium position at its first maximum extension.