Math Problem Statement

A 6.0 kg block initially at rest is pushed against a wall by a 100 N force as shown. The coefficient of kinetic friction is 0.30 while the coefficient of static friction is 0.50. What is true of the friction acting on the block after a time of 1 second?

Solution

The problem involves a block of mass 6.0kg6.0 \, \text{kg} being pushed against a wall with a force of 100N100 \, \text{N} at an angle of 6060^\circ. There is both static and kinetic friction involved, and we need to determine the friction acting on the block after 1 second.

Step 1: Break Down the Forces

  • Normal force: The force pushing the block against the wall comes from the horizontal component of the applied force: Fnormal=100N×cos(60)=100×0.5=50N.F_{\text{normal}} = 100 \, \text{N} \times \cos(60^\circ) = 100 \times 0.5 = 50 \, \text{N}. This normal force determines the frictional forces.

  • Gravitational force: The weight of the block is: Fgravity=mg=6.0kg×9.8m/s2=58.8N.F_{\text{gravity}} = mg = 6.0 \, \text{kg} \times 9.8 \, \text{m/s}^2 = 58.8 \, \text{N}. This force acts downward.

Step 2: Determine Static Friction Maximum

Static friction acts to prevent the block from sliding, and its maximum value is: fstatic, max=μsFnormal=0.50×50N=25N.f_{\text{static, max}} = \mu_s F_{\text{normal}} = 0.50 \times 50 \, \text{N} = 25 \, \text{N}. This is the maximum force static friction can exert. If the gravitational force exceeds this, the block will start to slide.

Step 3: Analyze the Situation

  • The downward gravitational force is 58.8N58.8 \, \text{N}, which is greater than the maximum static friction of 25N25 \, \text{N}.
  • This means that static friction cannot hold the block, and it will start sliding downward.

Step 4: Kinetic Friction Acting on the Block

Once the block begins to move, kinetic friction will act against the motion (upward, as the block is sliding downward). The kinetic friction is given by: fkinetic=μkFnormal=0.30×50N=15N.f_{\text{kinetic}} = \mu_k F_{\text{normal}} = 0.30 \times 50 \, \text{N} = 15 \, \text{N}. So, the kinetic friction acts upward as the block slides down the wall.

Conclusion:

The correct answer is:

  • Kinetic friction acts upward on the block.

Do you want further details or have any questions about this solution? Here are some related questions:

  1. How is the normal force affected if the angle of the applied force changes?
  2. What would happen if the coefficient of static friction were higher?
  3. How do you calculate the acceleration of the block once it starts sliding?
  4. What is the difference between static and kinetic friction in general?
  5. How would the friction change if the block were pushed upward at an angle instead of downward?

Tip: Always resolve forces into their perpendicular components to analyze problems involving friction and inclines!

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Math Problem Analysis

Mathematical Concepts

Friction
Forces
Newton's Laws

Formulas

Normal force: F_normal = F_applied * cos(θ)
Weight: F_gravity = mg
Static friction: f_static_max = μ_s * F_normal
Kinetic friction: f_kinetic = μ_k * F_normal

Theorems

Newton's Second Law
Friction Laws

Suitable Grade Level

Grades 10-12