Friction Force of a rubber ball

[I <3 Physic]

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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[RUber]

Why did you set ##\mu k = 1##?

 

[I <3 Physic]

that was the value that was given to me, on my coefficients of friction sheet.

 

[462chevelle]

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

 

[HalfLostAndSearching]

I would like to point out that there are a few things that need clarification in this scenario. First, the statement mentions that the ball is rolled across the floor with an acceleration of 4.0 m/s^2[E], but it does not specify the initial velocity of the ball or the distance over which it is rolled. These parameters are important in determining the final velocity and the time it takes for the ball to come to a stop.

Additionally, the value of µk (coefficient of kinetic friction) is given as 1, which is quite high for a rubber ball on a dry concrete floor. This value is usually lower and depends on various factors such as the type of rubber, surface roughness, and temperature. A more accurate value of µk would be needed to calculate the friction force accurately.

Assuming that the initial velocity is zero and the ball is rolled a short distance, we can use the equation Fnet = ma to calculate the friction force required to stop the ball. The net force acting on the ball in the horizontal direction is the friction force, which is equal to the product of µk and the normal force (Fn = mg). Therefore, we can write the equation as follows:

Fnet = Fk = µkFn = µkmg = ma

Solving for Fk, we get:

Fk = ma/µk = (2.5 kg)(4.0 m/s^2[E]) / 1 = 10 N[E]

This means that a friction force of 10 N[E] is required to stop the ball. However, as mentioned earlier, this value may vary depending on the actual value of µk and other factors. A more thorough analysis would be needed to accurately determine the friction force required to stop the ball in this scenario.