What Is the Correct Belt Velocity to Stop Boxes on an Incline with Friction?

• VSCCEGR
In summary: Solving for v, I get v = 3.87 m/sIn summary, the boxes are transported on a belt with a constant velocity, but experience friction with a coefficient of .4 when sliding down an inclined surface. By setting up a force diagram and using the equations for work and kinetic energy, it can be determined that the velocity of the belt must be 3.87 m/s in order for the boxes to stop at point B.
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:
Draw a force diagram. You know they are traveling 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.

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:
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?

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

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.

VSCCEGR said:
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?

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.

Marlon:

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

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

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

??

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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,

Last edited:
VSCCEGR said:
I have tried this:

N=mgcos(theta)
Ff=uN
W=Ff(d)
Ff(d)=.5mv^2
Your mistake was to ignore the change in potential energy in your energy equation:
$$F_f d = 1/2 m v^2 + mg d\sin\theta$$

Q: What is energy?

Energy is the capacity of a system to do work or produce heat. It can exist in various forms such as mechanical, thermal, electrical, chemical, and nuclear.

Q: What is friction?

Friction is a force that opposes motion between two surfaces in contact. It is caused by the roughness of the surfaces and can result in the conversion of kinetic energy into heat.

Q: How does friction affect energy?

Friction can decrease the amount of energy available for a system to do work. This is because some of the energy is converted into heat due to the resistance created by friction.

Q: How can friction be reduced?

Friction can be reduced by using lubricants, polishing surfaces to make them smoother, or using materials with lower coefficients of friction. Additionally, reducing the weight or pressure on the surfaces can also decrease friction.

Q: How is energy conserved in a system with friction?

Energy is conserved in a system with friction because although some of the energy is lost due to friction, the total amount of energy in the system remains constant. This is because energy cannot be created or destroyed, only transferred or converted from one form to another.

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