Relationship between coefficient of kinetic friction and theta?

- A block is projected up an inclined plane that makes an angle θ to the horizontal. It returnsto its initial position with half its initial speed. What is the coefficient of kinetic frictionbetween the block and the plane in terms of the angle of the incline?

So i obtained coefficient of kinetic friction, u=(5/3)tan(theta). Is this correct? If not can you guide me in the right direction?

Re: Relationship between coefficient of kinetic friction and theta?

Quote:

Originally Posted by

**turbozz** - A block is projected up an inclined plane that makes an angle θ to the horizontal. It returnsto its initial position with half its initial speed. What is the coefficient of kinetic frictionbetween the block and the plane in terms of the angle of the incline?

So i obtained coefficient of kinetic friction, u=(5/3)tan(theta). Is this correct? If not can you guide me in the right direction?

You've got the fraction flipped. It's (3/5) tan(theta).

Good job! That's what we call a "non-trivial" problem.

Would you care to share your solution so others might perhaps get some benefit from it?

-Dan

Re: Relationship between coefficient of kinetic friction and theta?

I used the kinetic energy-work theorem. That is, K(final) - K (initial)=W=F(net)(x(f)-x(i)).

And I used the theorem to describe motion on the way up and on the way down, and after simply solved the equations for the coefficient of k.friction.

I was wondering how you obtained 3/5tan(theta)? I keep obtaining 5/3tan(theta)??

Re: Relationship between coefficient of kinetic friction and theta?

In stages. For the motion of the block as a whole is to be propelled up the plane with a speed v_0 to a distance d along the plane and back down to its original point, where now has half the speed it originally did. So over the whole trip we have

$\displaystyle W_{nc} = \Delta E = \Delta T + \Delta U = \left ( \frac{1}{2} m \left ( \frac{v_0}{2} \right ) ^2 - \frac{1}{2} m v_0^2 \right )$

(We can neglect the gravitational PE since we are starting and ending at the same point.)

The non-conservative work is due to friction for which the frictional force is always directed opposite the direction of motion. So the friction force does $\displaystyle - 2 d f = -2 d \mu _k N$ of work, where d is the maximum distance up the plane, measured along the plane.

Combining this we have

$\displaystyle -2 d \mu _k N = \frac{1}{8} m v_0^2 - \frac{1}{2} m v_0^2 = - \frac{3}{8} m v_0 ^2$

We need a value for d. I did this as a Newton's 2nd Law problem. Skipping the details I get that

$\displaystyle N = mg~cos( \theta )$

$\displaystyle f = \mu _k mg ~cos(\theta)$

and the acceleration of the block up the incline

$\displaystyle a = \mu _k g ~ cos( \theta ) + g~sin( \theta )$

down the plane and it is a constant.

Putting this together with a simple kinematics problem I get that

$\displaystyle d = \frac{v_0^2}{2(g~sin( \theta ) + \mu _k g~cos(\theta ))}$

The energy equation now reads:

$\displaystyle -2 \left ( \frac{v_0^2}{2(g~sin( \theta ) + \mu _k g~cos(\theta ))} \left ) \mu _k ( mg~cos( \theta ) )= - \frac{3}{8} m v_0 ^2$

After a fair amount of simplifying I get

$\displaystyle \frac{\mu _k ~ cos(\theta)}{sin( \theta ) + \mu _k~cos( \theta )} = \frac{3}{8}$

Which, after the cross multiplication:

$\displaystyle 5 \mu _k~cos(\theta) = 3~sin(\theta)$

and finally:

$\displaystyle \mu _k = \frac{3}{5} ~ tan(\theta)$

I might have done this the hard way. I have a rather bad habit of doing that.

-Dan

Re: Relationship between coefficient of kinetic friction and theta?

My approach may be a bit simpler:

Since final velocity is 1/2 inital velocity the KE at the end is 1/4 the KE at the beginning. The loss of 3/4 of the initial KE is due to work performed by friction as the block travels up the ramp and back down again:

$\displaystyle (\frac 3 4 )( \frac 1 2 ) m v_i^2 = 2 \mu mg \cos (\theta) d $

$\displaystyle \frac 1 2 m v_i^2 = \frac 8 3 \mu mg \cos (\theta)d$

where d = length of the ramp.

When the block reaches its max eleveration its PE plus work done by friction = initial KE:

$\displaystyle mgd \sin (\theta) + mg \mu d \cos (\theta)= \frac 1 2 m v_i^2$

Combining the two equations:

$\displaystyle \frac 8 3 \mu mg \cos (\theta)d = mgd \sin (\theta) + mg \mu d \cos (\theta$

$\displaystyle \frac 5 3 \mu \cos \theta = \sin \theta $

$\displaystyle \mu = \frac 3 5 \tan \theta $

Re: Relationship between coefficient of kinetic friction and theta?

Quote:

Originally Posted by

**topsquark** I might have done this the hard way. I have a rather bad habit of doing that.

-Dan

Told you!

-Dan

Re: Relationship between coefficient of kinetic friction and theta?

Quote:

Originally Posted by

**topsquark**

Putting this together with a simple kinematics problem I get that

$\displaystyle d = \frac{v_0^2}{2(g~sin( \theta ) + \mu _k g~cos(\theta ))}$

How do you obtain this? I would think v^2 - v_0^2 =2ad, but then why would v become 0 in this case?

Re: Relationship between coefficient of kinetic friction and theta?

Quote:

Originally Posted by

**turbozz** How do you obtain this? I would think v^2 - v_0^2 =2ad, but then why would v become 0 in this case?

One of the things I glossed over a bit on in my explanation. To find d I have used the part of the problem where the block is projected up the incline and goes to it's highest position a distance d up the slope and momentarily stops (the turning point.) v = 0 m/s there. I did it as a kinematics problem, but you could easily do it as an energy problem as well.

-Dan