A 5 kg block is being pulled to the right with a 20 N force. This causes the block to slide along the table with an acceleration of 2 m/s2. What is the coefficient of kinetic friction between the block and table?
F = m a
20 - mu (5)(9.8) = 5 (2)
solve for mu
To find the coefficient of kinetic friction between the block and the table, we need to use Newton's second law of motion and the formula for friction.
1. Start with Newton's second law of motion, which states that Force (F) equals mass (m) multiplied by acceleration (a):
F = m * a
2. In this case, the force exerted on the block is the applied force minus the friction force. So we have:
F_applied - F_friction = m * a
3. Rearrange the equation to solve for the friction force (F_friction):
F_friction = F_applied - m * a
4. The friction force can be written as the coefficient of kinetic friction (μ) multiplied by the normal force (F_normal):
F_friction = μ * F_normal
5. The normal force is equal to the weight of the block, which is the mass (m) multiplied by the acceleration due to gravity (g):
F_normal = m * g
6. Combine equations 4 and 5 to express the friction force in terms of μ:
F_friction = μ * m * g
7. Substitute equation 6 into equation 3:
μ * m * g = F_applied - m * a
8. Rearrange the equation to solve for the coefficient of kinetic friction (μ):
μ = (F_applied - m * a) / (m * g)
9. Plug in the given values:
μ = (20 N - (5 kg * 2 m/s^2)) / (5 kg * 9.8 m/s^2)
10. Calculate the coefficient of kinetic friction (μ) using the equation:
μ = (20 N - 10 N) / 49 N
11. Simplify the equation:
μ = 10 N / 49 N
12. Calculate the final answer:
μ ≈ 0.204
Therefore, the coefficient of kinetic friction between the block and table is approximately 0.204.