grr, some physics help

[blike]

I missed a rather simple problem on my physics test, and I'm not quite sure what I did wrong. I'm sure its a stupid mistake somewhere, be it in the formulas, calculations, or reasoning.

 

A motor is pulling a block up a 50m slope inclined at 20' above the horizontal at a constant speed of 2.5m/s. The coefficient of kinetic friction is 0.06. Find the power in the motor required to do this if the block has a mass of 100-kg.

 

W = f:delta:x

Power = W/:delta:t

f_kinetic = :mu:n

n = mgcos:theta:

 

:theta: = 20'

:mu:= 0.06

m = 100kg

a = 0

v_o = 2.5 m/s

:delta:x = 50m

 

Because there is no acceleration, the net force on the x-axis is 0

 

:sum:F_x = F_motor - f_kinetic = 0

 

Thus, F_motor = f_kinetic

 

f_kinetic = .06 * mgcos(20)

f_kinetic = 55.25N

 

F_motor = 55.25N

 

Plugging this into the "work" equation: w=F:delta:x

 

W = (55.25N)(50m)

W =2762.70J

 

Plugging this into power: P = W/:delta:t

 

P = 2762.70J / 20s = 138.135

 

Where did I go wrong?

[VendingMenace]

perhaps this is the answer....

sorry i don't know how to do the subscript stuff, so this may be somewhat hard to read

 

i think you went wrong when you calculated the the work.

 

As you rightly stated, there is no work in the X direction. Thus, all work MUST be in the y direction. IN other words, all the work that is done is expended to lift the block off the ground. SO you must figure out how high the block is lifted. THis is given by the equation;

 

hieghth = (length of path) * sin(theta)

 

IN this case it is;

 

hieght = 50m * sin(20)

 

 

Thus, your work equation for this problem should look like;

 

W = F * (delta X) -- where we are only considered with change in the y coordinate

 

W = (55.25N)(17.1m)

W = 944.8J

 

thus, power would be...

 

P = (944.8J)/(20s)

= 47.24w

 

 

well, that is the only problem that i saw in your reasoning. Hope that helps some

[Sayonara]

The tags for _subscript and ^superscript are sub and sup respectively.

 

Quote this post to see them

[swansont]

You want the total, not just in the x-direction.

 

Since the net acceleration is zero, you can get the power with P=Fv, and the net force is the vector sum of the weight, friction, normal, and rope. You want to solve for the force from the rope.

 

Alternately, you can solve for the work done by friction and potential energy in a given amount of time, since you have a constant known speed. Work/time = P

[blike]

VendingMenace, thanks for your response!

 

I used a coordinate system in which the y-axis runs parallel with the normal force, and thus both the horizontal and vertical force are summed together.

 

As you rightly stated, there is no work in the X direction

 

I would argue that the motor must maintain a constant tension (on the rope holding the block) of 55.25N on the block for a total of 50m, regardless of the direction of motion [i.e. coordinate system doesn't really mater, as long as the calculations are adjusted appropriately]. Thus, while the net force on the x-vector of the block is 0, the motor must produce a force of 55.25N in order to haul it up the incline.

 

 

swansont, I would think that the y-components of the forces do not matter, because the work is being done by gravity, not the motor.

[JaKiri]

Since when does gravity help one to pull a block up a slope?

[Sayonara]

lol

 

Aye, there's the rub.

[blike]

MrL_JaKiri said in post #6 :

Since when does gravity help one to pull a block up a slope?

 

The force of gravity is already in the problem in the form of "force of kinetic friction", which is the coefficient of kinetic friction * the normal force. The normal force is mgcos:theta:, which is the force the box exerts on the object due to gravity.

[blike]

I think what's confusing everyone is that I've shifted the coordinates so that the slope of the incline is parallel with the x-axis, so as the only motion the block has is in the x-direction.

[swansont]

blike said in post #8 :

 

 

The force of gravity is already in the problem in the form of "force of kinetic friction", which is the coefficient of kinetic friction * the normal force. The normal force is mgcos:theta:, which is the force the box exerts on the object due to gravity.

 

No, because that term would go away if the coefficient of friction went to zero, and you still have gravitational potential energy for which you must account.

[swansont]

blike said in post #5 :

swansont, I would think that the y-components of the forces do not matter, because the work is being done by gravity, not the motor.

 

The question states that you are to find the power of the motor, which is related to the work it has to do.

[blike]

I believe I found my mistake.

 

:sum:F_x = F_motor - f_kinetic = 0

 

I forgot to include mgsin:theta: which is probably what you have been saying..

 

Thus, the proper sum for the x-vector forces is:

 

:sum:F_x = Force_motor - f_kinetic - mgsin:theta:

 

Then,

 

Force_motor = f_kinetic + mgsin:theta:

 

I was neglecting gravity's affect on the box, thinking that it was already factored into the force of kinetic friction.

 

swanson, you the man.

 

thanks everyone else!

[aman]

Thanks, that was fun to follow. Wish I coulda jumped in but I was to busy digesting it as it came along. I'm slow but I still can get there. Makes sense it is the total of friction and gravity the motor overcomes.

Just aman