Hi, the question is: a object placed on a rough floor has v_{0 }= 10 m/s and slides to rest after 25m. Find μ. D = distance.

The notes I have are 1/2 mv^{2 }= Ffr.D.

Then, 1/2 mv^{2 }= μ.Fn.D,

1/2 mv^{2 }= μ.m.g.D

First is F.d in these instances a dot product or multiplication?

Second, the above formula doesn't make logical sense to me, but are both equal to the work done. Are they the same work done? (I'm guessing that the 1/2 mv^{2 }is Kinetic final - Kinetic intial = work. Right?

How are the forces of friction over distance the same as the kinetic energy?

Please asist further.

Stu
Jul 16, 2014

#1**+10 **

The first equation, which should probably be written as 1/2*m*v_{0}^{2} = F_{fr}*D, says that the initial kinetic energy is equal to the *work done* by friction in bringing the object to rest (the right-hand side is *frictional force*distance *which is work, which is a form of energy transfer).

The frictional force is usually represented by F_{fr} = μ*F_{n}, where F_{n} is the normal force; that is, the force holding the object on the floor, namely, its weight (so F_{n} = m*g, where m is the object's mass and g is the acceleration of gravity). μ is called the coefficient of friction.

So, putting this together, we end up with: 1/2*m*v_{0}^{2} = μ*m*g*D which can be rearranged to find μ = (1/2*m*v_{0}^{2})/(m*g*D) and the m cancels so μ = (1/2*v_{0}^{2})/(g*D).

Does this make it clearer?

Alan
Jul 16, 2014

#1**+10 **

Best Answer

The first equation, which should probably be written as 1/2*m*v_{0}^{2} = F_{fr}*D, says that the initial kinetic energy is equal to the *work done* by friction in bringing the object to rest (the right-hand side is *frictional force*distance *which is work, which is a form of energy transfer).

The frictional force is usually represented by F_{fr} = μ*F_{n}, where F_{n} is the normal force; that is, the force holding the object on the floor, namely, its weight (so F_{n} = m*g, where m is the object's mass and g is the acceleration of gravity). μ is called the coefficient of friction.

So, putting this together, we end up with: 1/2*m*v_{0}^{2} = μ*m*g*D which can be rearranged to find μ = (1/2*m*v_{0}^{2})/(m*g*D) and the m cancels so μ = (1/2*v_{0}^{2})/(g*D).

Does this make it clearer?

Alan
Jul 16, 2014

#2**0 **

Yes it does thank you. Can you also assist with the question: a 2kg block is projected up a plane of angle 30 degrees with the horizontal. V_{0} is 5m/s and μ = 0. Find the distance the lock travels along the incline.

Please explain process of forces acting on it, as you did in the last post @Alan. I want to see how to reason logically and understand the process. Thanks.

Stu
Jul 16, 2014

#3**+5 **

I guess I should note that this is the same as using kinetic energy = force*distance, but now you must use total force along the line of motion, not just the frictional force (in the case of the object on the horizontal floor the frictional force *was* the total force along the line of motion).

Alan
Jul 16, 2014

#4**0 **

Alan, thanks. I got 2.55 m not rounded and the answers given had 2.6 m rounded up. How I got that so you can check where I went wrong was,

1/2 mv_{0}^2 = -mg sin(theta)*d

Cancle m, transpose to make d the subject.

Therefore, 1*v_{0}^2/2*g*sin(theta) =d.

And with values it should come out.

Two questions come to mind, why is it -mgsin(theta)?

Also, why did you take the line with calculating the distance with displacement equation?

Which is the right answer from my answer and yours?

Stu
Jul 16, 2014

#5**0 **

2.55 is correct if you use 9.8 for g and zero for μ. I used 9.81 for g and 0.1 for μ which is why our final numerical results are slightly different.

If you use the kinetic energy approach you don't need the negative sign on the m*g*sin(30). The sign in my approach indicates that the direction *down* the slope is negative, and that the acceleration *up* the slope is negative (i.e. it is actually a *deceleration*).

Alan
Jul 16, 2014