Final answer:
The force of kinetic friction acting on a 20.0-kg rock with a coefficient of kinetic friction of 0.200 on a horizontal surface is calculated using the normal force and the formula λk = μ_k × N. The result is a force of 39.2 N.
Step-by-step explanation:
The student has asked about the force of kinetic friction acting on a 20.0-kg rock that was sliding on a rough, horizontal surface and has come to a stop due to friction. The coefficient of kinetic friction (μ_k) is given as 0.200.
To find the force of kinetic friction (λk), we can use the formula:
λk = μ_k × N
where N is the normal force, which is equal to the weight of the rock if the surface is horizontal and there are no other forces acting in the vertical direction. Since weight (W) can be calculated as the mass (m) times the acceleration due to gravity (g), we have:
W = m × g
where:
- m = 20.0 kg (mass of the rock)
- g = 9.8 m/s² (acceleration due to gravity)
So, the normal force (N) is:
N = W = 20.0 kg × 9.8 m/s² = 196 N
Now, we can find the force of kinetic friction (λk):
λk = 0.200 × 196 N = 39.2 N
Hence, the force of kinetic friction acting on the rock is 39.2 N.