Newton's 2nd Law and a rolling ball

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SUMMARY

The discussion focuses on two physics problems involving Newton's 2nd Law. The first problem involves a soccer ball rolling down a hill with a final speed of 10.0 m/s and a coefficient of friction of 0.80. The key to solving this problem is determining the friction force on the level surface, which is necessary to calculate the ball's deceleration and distance traveled before coming to rest. The second problem involves a pig sliding down a 35-degree incline, where the coefficient of kinetic friction must be calculated based on the difference in acceleration between a frictionless incline and one with friction. The acceleration on the incline with friction is less due to the opposing friction force.

PREREQUISITES
  • Understanding of Newton's 2nd Law of Motion
  • Knowledge of friction coefficients and their implications
  • Ability to apply kinematic equations for motion analysis
  • Familiarity with vector components of forces
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  • Calculate the deceleration of a rolling object using friction coefficients
  • Explore kinematic equations for motion on inclined planes
  • Learn about the effects of friction on acceleration in physics
  • Study the relationship between force, mass, and acceleration in various contexts
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Students studying physics, educators teaching mechanics, and anyone interested in applying Newton's laws to real-world scenarios involving motion and friction.

alikat785
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Hello!
I have a few homework problems that I'm just not sure what to do with:

1) A soccer ball rolls down a hill, reaching the bottom with a speed of 10.0 m/s. The coefficient of friction between the ball and the grass on the hill is .80. How far does the ball travel on the surface before coming to rest?

I know that the initial velocity is 0 m/s, and that the final velocity is 10 m/s, but I'm not sure what to do to find the answer.

2) A pig slides down a 35 degree incline in twice the time it would take to slide down a frictionless 35 degree incline. What is the coefficient of kinetic friction between the pig and the incline?

I found the acceleration of the pig on a frictionless incline using the equation a=9.8 sin 35 = 5.62 m/s^2. I thought maybe I could put it into the equation d=Vi*t + .5*a*t^2? so that

Vi*t + .5*(5.62)*t^2 = Vi*2t +.5(a)(2t)^2

since the distance the pig travels is the same for both inclines, and I think the initial velocity would be 0? But I'm not sure what the acceleration for the incline with friction would be- the same or different than the one without friction?

I would really appreciate any help anyone can give me. Thanks in advance!
Alison
 
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Well when the pig slides and there's friction,
the forces on the x-axis are

mgsin\theta - \mu mgcos\theta = ma

g(sin\theta - \mu cos\theta) = a
 
This first problem "1) A soccer ball rolls down a hill, reaching the bottom with a speed of 10.0 m/s. The coefficient of friction between the ball and the grass on the hill is .80. How far does the ball travel on the surface before coming to rest?" is slightly ambiguous. It says the friction between the ball and the grass on the hill is 0.8 but what you really need is the friction between the ball and the level surface. I presume that is what is actually meant. The motion on the hill is irrelevant. You know its speed at the bottom of the hill is 10 m/s. If you know the horizontal (friction) force on the ball you can calculate it deceleration and then use the usual acceleration-velocity-distance formulas to calculate how far it travels. There's still an ambiguity: normally, for object moving over a surface, "coefficient of friction" means "friction force equals coefficient of friction times weight" but you are not given the weight (or mass). In problems involving motion through water or air, "coefficient of friction" might mean "friction force equal coefficient of friction times speed" but I don't think that's intended here.

As for the "sliding pig" problem, "But I'm not sure what the acceleration for the incline with friction would be- the same or different than the one without friction?" That's the whole point. F= ma. The force, without friction, is just gravity. WITH friction, there is also a friction force so certainly the acceleration is different! The friction force is directed back up the slope so, with friction, the total force and acceleration will be smaller.
Caution: the gravity force is directed straight down. The friction force is parallel to the slope. You will need to add the components of the forces.
 

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