A block of mass m = 3.45 kg slides along a horizontal table with speed v0 = 2.00 m/s. At x = 0 it hits a spring with spring constant k = 80.00 N/m and it also begins to experience a friction force. The coefficient of friction is given by ¦Ì = 0.250. How far has the spring compressed by the time the block first momentarily comes to rest?

To find the distance the spring has compressed when the block first momentarily comes to rest, we need to calculate the total work done on the block. This work is equal to the work done by the spring and the work done by friction.

The work done by the spring can be calculated using Hooke's Law:

Work_spring = (1/2)kx^2

where k is the spring constant and x is the distance the spring is compressed.

To find the distance x, we can use the equation for potential energy stored in the spring:

(1/2)kx^2 = (1/2)mv0^2

where m is the mass of the block and v0 is the initial velocity.

Rearranging this equation, we get:

x^2 = (mv0^2) / k

x = sqrt((mv0^2) / k)

Now, let's calculate the values:

m = 3.45 kg
v0 = 2.00 m/s
k = 80.00 N/m

x = sqrt((3.45 kg * (2.00 m/s)^2) / 80.00 N/m)

x = sqrt((3.45 kg * 4.00 m^2/s^2) / 80.00 N/m)

x = sqrt(0.1725 kg * m^2/s^2 / N/m)

x = sqrt(0.1725 kg * m)

x = 0.415 m

Therefore, the spring has compressed by a distance of 0.415 meters when the block first momentarily comes to rest.