The coefficient of kinetic friction (μ_k) between two surfaces is a measure of how much frictional force opposes the motion of an object sliding on a surface. In this case, we are given that the coefficient of kinetic friction between the block and the sand-covered ice is 0.28.
To find the frictional force acting on the block when it slides onto the sand-covered ice, we can use the following formula:
\[ F_f = \mu_k \cdot F_N \]
where:
- \( F_f \) is the frictional force,
- \( \mu_k \) is the coefficient of kinetic friction,
- \( F_N \) is the normal force.
For an object resting on a horizontal surface, the normal force (F_N) is equal to the weight of the object (mg), where:
- \( m = 20 \text{ kg} \) (mass of the block),
- \( g = 9.81 \text{ m/s}^2 \) (acceleration due to gravity).
The normal force can be calculated as:
\[ F_N = m \cdot g = 20 , \text{kg} \cdot 9.81 , \text{m/s}^2 = 196.2 , \text{N} \]
Now, substituting the values into the equation for frictional force:
\[ F_f = \mu_k \cdot F_N = 0.28 \cdot 196.2 , \text{N} \]
Calculating \( F_f \):
\[ F_f = 0.28 \cdot 196.2 \approx 54.97 , \text{N} \]
So, while the frictional force opposing the motion of the block when it slides onto the sand-covered ice is approximately 54.97 N, the coefficient of kinetic friction has already been provided as 0.28. This coefficient describes the ratio of the frictional force to the normal force and is the quantity you were asked to calculate.
To summarize:
- The coefficient of kinetic friction between the block and the sand-covered ice is 0.28.
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