Math Problem Statement

A block moves along a smooth curved path and then transitions onto a rough inclined plane with a coefficient of kinetic friction μ_k = 0.2, stopping at a height y_maks. What is the equation for the final height y_maks?

Solution

The problem involves a block moving along a smooth curved path, then transitioning to a rough inclined plane where kinetic friction, denoted by μk=0.2\mu_k = 0.2, slows it down until it stops at a height ymaksy_{\text{maks}}, as shown in the image.

The question is asking for the equation that represents the final position ymaksy_{\text{maks}} of the object after stopping on the inclined plane.

Key Points:

  • Kinetic Friction: μk=0.2\mu_k = 0.2
  • Inclined plane angle: θ\theta
  • Initial height: hh
  • Energy considerations: The initial energy is purely potential, and the energy loss is due to friction as the object moves up the inclined plane.

Solving the Problem:

  1. Initial Energy:

    • The object starts from rest at height hh, so its potential energy is PE=mghPE = mgh.
  2. Energy Loss due to Friction:

    • As the object moves up the incline, friction does work against the motion.
    • The work done by friction is Wf=μkmgcosθdW_f = \mu_k mg \cos \theta \cdot d, where dd is the distance traveled up the incline.
  3. Final Position:

    • At the maximum height ymaksy_{\text{maks}}, the object stops, meaning all of the remaining energy is potential energy at ymaksy_{\text{maks}}, which is PE=mgymaksPE = mg y_{\text{maks}}.
    • Using the principle of conservation of energy: Initial potential energy = Final potential energy + Work done by friction.

    The equation becomes: mgh=mgymaks+μkmgcosθdmgh = mg y_{\text{maks}} + \mu_k mg \cos \theta \cdot d where dd can be related to ymaksy_{\text{maks}} through the incline's geometry: d=ymakssinθd = \frac{y_{\text{maks}}}{\sin \theta}.

  4. Solving for ymaksy_{\text{maks}}: After substituting dd and simplifying, we obtain the expression for ymaksy_{\text{maks}}: ymaks=h1+μkcotθy_{\text{maks}} = \frac{h}{1 + \mu_k \cot \theta}

Conclusion:

Based on the above analysis, the correct option is:

  • Answer: D.

Would you like further clarification or a deeper breakdown of any part of the solution?

Here are 5 related questions to explore the topic:

  1. How does the angle of incline affect the final height ymaksy_{\text{maks}}?
  2. What role does the coefficient of friction μk\mu_k play in determining the stopping point of the object?
  3. How would the result change if there were no friction on the inclined plane?
  4. What if the object were to start with some initial kinetic energy? How would the equation change?
  5. How do we derive the relationship between dd and ymaksy_{\text{maks}} using trigonometry?

Tip: When dealing with energy problems, always account for both potential and kinetic energy, and consider the work done by external forces like friction.

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Math Problem Analysis

Mathematical Concepts

Kinetic Friction
Energy Conservation
Inclined Plane Geometry

Formulas

Potential energy: PE = mgh
Work done by friction: W_f = μ_k mg cos(θ) · d
Distance: d = y_maks / sin(θ)
Energy conservation: mgh = mgy_maks + μ_k mg cos(θ) · d

Theorems

Conservation of Energy

Suitable Grade Level

Grades 11-12 (High School Physics)