The block will travel approximately 1.97 meters before coming to rest.
To find the distance the block will travel before coming to rest, we can use the work-energy principle. The work done by the external force (the initial kick) is equal to the work done against friction, which is converted into kinetic energy and then dissipated.
The work done by the external force is given by:
\[ W_{\text{external}} = \frac{1}{2} m v_{\text{initial}}^2 \]
The work done against friction is given by:
\[ W_{\text{friction}} = \mu_k m g d \]
where:
- \( m \) is the mass of the block (2.3 kg),
- \( v_{\text{initial}} \) is the initial velocity of the block (5.1 m/s),
- \( \mu_k \) is the coefficient of kinetic friction (0.67),
- \( g \) is the acceleration due to gravity (approximately 9.8 m/s²),
- \( d \) is the distance traveled.
The work-energy principle states that the work done by external forces is equal to the change in kinetic energy:
\[ W_{\text{external}} = \Delta KE \]
Since the block comes to rest, the final kinetic energy (\( KE_{\text{final}} \)) is zero:
\[ \frac{1}{2} m v_{\text{initial}}^2 = \mu_k m g d \]
Now, solve for \( d \):
\[ d = \frac{\frac{1}{2} m v_{\text{initial}}^2}{\mu_k m g} \]
\[ d = \frac{v_{\text{initial}}^2}{2 \mu_k g} \]
Plug in the known values:
\[ d = \frac{(5.1 \, \text{m/s})^2}{2 \times 0.67 \times 9.8 \, \text{m/s}^2} \]
\[ d \approx \frac{26.01}{13.2} \]
\[ d \approx 1.97 \, \text{m} \]
Therefore, the block will travel approximately 1.97 meters before coming to rest.