# Energy & Friction Problem Help

1. Apr 5, 2005

### VSCCEGR

Boxes are transported by a belt w/a vel. Vo to a fixed incline @ A where they slide & fall off @ B. Knowing that the coefficent of kinetic friction is u=.4 Determine the Vel. of the belt if the boxes are to have a Vel. of 0 and stop @ B.

There is a constant velocity while the boxes are on the belt, but from A to B there is friction of .4. AB is not a part of the belt, but a 15* inclined surfface.

I have tried this:

N=mgcos(theta)
Ff=uN
W=Ff(d)
Ff(d)=.5mv^2

This Gives 6.73m/s

Last edited: Apr 5, 2005
2. Apr 5, 2005

### whozum

Draw a force diagram. You know they are travelling with cosntant velocity, so the net force in every direction is zero. This should be fairly easy.

Find the normal force, and given the friction coefficient, you can find the force of friction.

I can't decipher from your image whether the area labeled 6 meters is an area of freefall or part of the belt.

3. Apr 5, 2005

### whozum

Anyway, The boxes have a certain kinetic energy imparted to them by the belt:

$$KE_{box} = \frac{1}{2} mv_0^2$$

This energy will be completely dissipated by friction once the boxes reach point B, a distance of 6 meters away. You can find the force of friction by knowing that it dissipates the energy of the box in 6m, or:

The change in energy is the total kinetic energy of the boxes, $\Delta W = KE$

$$F = \frac{\Delta W}{d} = \frac{mv_0^2}{12}$$

$$F = \mu_k N = 0.4 N$$ and $$N = mgcos(15) = 9.46m$$ so:

$$(0.4)(9.46m) = \frac{mv_0^2}{12}$$

Cancel the masses, and you should get ~6 m/s

*well if the answer is 3.87 then somethings up.

Last edited: Apr 5, 2005
4. Apr 5, 2005

### VSCCEGR

W being work

Then W=KE2-KE1 not KE=Delta.W

KE2 Should=0 because the box will be at rest. therefor, W=KE1
KE1=.5mv^2 and W=Fd
F is where I think I am going wrong.
the Ff (friction force) isn't the only force acting is it?

5. Apr 5, 2005

### marlon

But along the direction of the incline, you also have a force component coming from gravity, which is equal to
-mgsin(theta)...

So the total net force should be $$-mgsin(15) + 0.4mgcos(15)$$

then proceed like whozum explained and you will get the 3.87m/s

marlon

6. Apr 5, 2005

### whozum

I was thinking this, but I figured that gravity would be negated by the normal force since it IS on the plane, and the vertical component of the normal force eliminated its effects. Now that I'm thinking about it that was what it is.

7. Apr 5, 2005

### whozum

Were explaining the same thing with different terminology. I was explaining to you why I changed from W to KE in the next equation, the change in work will be equivalent to the Ke_f-Ke_i, since Ke_i = 0, then change in work will be Ke_f. Change in work is referring to the change in energy.

I just say work too much.

8. Apr 5, 2005

### whozum

Marlon:

$$F_{net} = mgcos(15)-mgsin(15) = 6.92m$$

$$F_{net} = \frac{mv^2}{12}$$

$$v_0 = \sqrt{\frac{12F_{net}}{m}} = \sqrt{\frac{12(6.92m)} {m} } = 9.11m/s$$

??

Last edited: Apr 5, 2005
9. Apr 5, 2005

### VSCCEGR

PRAISE U GUYS!

sum F=1.25m
F(6)=.5mV^2 (Why are ya'll dividing by 12????)
7.5m=.5mV^2
15=V^2
V=3.87m/s

Thanks,
$$F_f d = 1/2 m v^2 + mg d\sin\theta$$