When an object slides at constant speed down an inclined plane, the coefficient of fraction may be approximately:
Correct answer: C. Tan θ
- A. Sin θ
- B. Cos θ
- C. Tan θ
- D. Cot θ
Explanation
The coefficient of friction is a measure of the frictional force between the object and the surface it is sliding on. It depends on the nature of the surfaces in contact and is typically denoted by the symbol 'μ'. The force of friction is given by the equation: f = μN where f is the force of friction, N is the normal force (the component of the weight of the object that is perpendicular to the inclined surface), and μ is the coefficient of friction.In the case where the object is sliding at constant speed down an inclined plane, the force of friction is equal in magnitude to the component of the weight of the object that is parallel to the plane. This component is given by: mg sin θ θ is the angle of inclination of the plane. Thus, we have: f = mg sin θ Substituting the equation for the force of friction into the equation for the coefficient of friction, we get: μ = f/N = (mg sin θ)/(mg cos θ) = tan θ Therefore, the coefficient of friction is approximately equal to the tangent of the angle of inclination of the plane. So, the correct option is (C) tan theta.
This question appeared on the ETEA MDCAT 2009 paper, which you can sit online with every answer explained.
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