What is the coefficient of kinetic friction µK of the surface

What is the coefficient of kinetic friction µK of the surface?

The correct answer and explanation is:

To find the coefficient of kinetic friction (µₖ) of a surface, we typically need information about the forces acting on an object sliding on that surface. The coefficient of kinetic friction is a dimensionless number that represents the ratio of the frictional force resisting the motion of two surfaces sliding past each other to the normal force pressing the two surfaces together.

Common formula for µₖ:

μk=FfrictionFnormal\mu_k = \frac{F_{\text{friction}}}{F_{\text{normal}}}

Where:

• FfrictionF_{\text{friction}} is the kinetic friction force (force opposing motion),
• FnormalF_{\text{normal}} is the normal force (usually the weight component perpendicular to the surface).

How to find µₖ in typical problems:

1. Using forces on a horizontal surface:

If a block of mass mm slides at a constant velocity on a horizontal surface, the kinetic friction force equals the applied force FF pulling it (because acceleration is zero).

• The normal force is Fnormal=mgF_{\text{normal}} = mg (gravity times mass).
• Friction force equals the applied force to maintain constant velocity.

Then, μk=Fmg\mu_k = \frac{F}{mg}

2. Using acceleration on a horizontal surface:

If the block accelerates, friction is less than the applied force. You can find friction from Newton’s second law: Fnet=ma=F−FfrictionF_{\text{net}} = ma = F – F_{\text{friction}}

Rearranged: Ffriction=F−maF_{\text{friction}} = F – ma

Then, μk=Ffrictionmg=F−mamg\mu_k = \frac{F_{\text{friction}}}{mg} = \frac{F – ma}{mg}

3. Using an inclined plane:

If the block slides down an incline at constant velocity or accelerating, the forces along the incline include gravity components and friction:

• Weight component parallel to incline: mgsin⁡θmg \sin \theta
• Weight component perpendicular: mgcos⁡θmg \cos \theta
• Kinetic friction: Ffriction=μkmgcos⁡θF_{\text{friction}} = \mu_k mg \cos \theta

If sliding at constant velocity: mgsin⁡θ=μkmgcos⁡θmg \sin \theta = \mu_k mg \cos \theta μk=tan⁡θ\mu_k = \tan \theta

If accelerating, Newton’s second law is applied to find friction first, then µₖ.

Summary:

The coefficient of kinetic friction μk\mu_k is a ratio of frictional force to normal force, indicating how “rough” or “slippery” the surface is during motion. It is always less than the coefficient of static friction for the same materials because moving surfaces typically require less force to keep sliding than to initiate movement.

Example:

Suppose a 10 kg block is pulled across a horizontal surface with a force of 30 N and accelerates at 1.5 m/s².

• Weight: Fnormal=mg=10×9.8=98 NF_{\text{normal}} = mg = 10 \times 9.8 = 98 \, N
• Net force: Fnet=ma=10×1.5=15 NF_{\text{net}} = ma = 10 \times 1.5 = 15 \, N
• Friction force: Ffriction=F−Fnet=30−15=15 NF_{\text{friction}} = F – F_{\text{net}} = 30 – 15 = 15 \, N
• Coefficient: μk=1598≈0.153\mu_k = \frac{15}{98} \approx 0.153