Friction Force of a rubber ball

AI Thread Summary
To determine the friction force required to stop a 2.5 kg rubber ball accelerating at 4.0 m/s² on concrete, the normal force (Fn) is calculated as 24.5 N using Fn = mg. The friction force (Fk) is then found to be 24.5 N, assuming a coefficient of kinetic friction (µk) of 1, which is considered unusually high for rubber on concrete. A participant points out that the typical coefficient of friction for rubber on concrete is around 0.8, suggesting the initial assumption may be incorrect. The discussion highlights confusion regarding the ball's movement despite the calculated forces, indicating a need for clarification on friction values. Accurate coefficients are essential for realistic physics problems.
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Homework Statement


If a 2.5 kg rubber ball is rolled across the dry concrete floor with an acceleration of 4.0 m/s^2[E]. Determine the friction force required to stop the ball. (Drawing a free body diagram will help illustrate this situation)

Homework Equations


Fk= µk(Fn)
F=ma

The Attempt at a Solution



a=4.0 m/s^2[E]

m=2.5kg
µk= 1Find Fn=mg

=2.5(9.81)

=24.5

Find friction force to stop the ballFk=1(24.5)

Fk=24.5

Find Fnet=ma

=2.5(4.0)

=10 N(E)
These numbers are telling me that the object should not be moving in the first place? Which is confusing because i am given an acceleration

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Why did you set ##\mu k = 1##?
 
that was the value that was given to me, on my coefficients of friction sheet.
 
That is very high, that is a friction i would expect to be static. and i thought rubber to concrete was somewhere around .8 mus
 

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