What Is the Coefficient of Friction for a Dry Pig on a 27-Degree Slide?

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The problem involves calculating the coefficient of friction for a dry pig sliding down a 27-degree slide, given that a greased pig slides down in half the time. To find the coefficient of friction (mu), apply Newton's second law to both scenarios, considering the difference in friction forces. The coefficient is a function of the angle of the slide, which is 27 degrees in this case. Understanding the relationship between time, friction, and angle is crucial for solving the problem. Breaking it down into smaller parts and seeking assistance can aid in grasping the concepts involved.
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Heres my problem..i just don't get it..and I am in AP Phys... :confused: :confused: :confused: :confused: :confused:

A greased pig slides down a slide in 1/2 the time it takes for the same pig to slide down the same slide "dry." What is the coefficient of friction of the "dry" pig as a function of (theta). If the angle the slide makes with the horizontal is 27(degrees) what is (Mu)??
 
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Apply Newton's 2nd Law on both cases, then use a kinematics formula with time and work it from there.
 


It's completely understandable to feel confused about a problem in AP Physics. It can be a difficult subject, and it's normal to struggle with certain concepts or problems. However, don't let that discourage you. Keep working at it and seeking help when needed. As for this specific problem, the first step is to understand the given information. A greased pig sliding down a slide in half the time as a dry pig means that the friction force is different in both cases. The coefficient of friction (mu) is a measure of the friction force between two surfaces. In this case, it would be a function of the angle (theta) of the slide. The angle (theta) given in the problem is 27 degrees, and using that, you can calculate the coefficient of friction. Remember to use the correct formula and units for (mu) and (theta). If you're still having trouble, try breaking down the problem into smaller parts and asking for help from your teacher or classmates. Keep practicing and don't get discouraged, you'll get it eventually!
 
Thread 'Variable mass system : water sprayed into a moving container'

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Starting with the mass considerations #m(t)# is mass of water #M_{c}# mass of container and #M(t)# mass of total system $$M(t) = M_{C} + m(t)$$ $$\Rightarrow \frac{dM(t)}{dt} = \frac{dm(t)}{dt}$$ $$P_i = Mv + u \, dm$$ $$P_f = (M + dm)(v + dv)$$ $$\Delta P = M \, dv + (v - u) \, dm$$ $$F = \frac{dP}{dt} = M \frac{dv}{dt} + (v - u) \frac{dm}{dt}$$ $$F = u \frac{dm}{dt} = \rho A u^2$$ from conservation of momentum , the cannon recoils with the same force which it applies. $$\quad \frac{dm}{dt}...
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