Who is right - me or the book - work and kinetic energy


by Hemmelig
Tags: book, energy, kinetic, work
Hemmelig
Hemmelig is offline
#1
Apr21-08, 10:43 AM
P: 22
My educated guess would be that i'm wrong

1. The problem statement, all variables and given/known data

A factory worker pushes a 30kg crate a distance of 4,5m along a floor at constant velocity, the coefficient of kinetic friction between the crate and floor is 0.25

The angle he pushes the crate with is 30 degrees below the horizontal



2. Relevant equations

Work = F * s * cos(angle)

Normal force = mg

3. The attempt at a solution

What i do know is that since the velocity is constant, the total work done on the crate is 0

So Work total =Wt = Work done by worker + Work done on crate by friction = 0

So Work done by worker = - work done on crate by friction

Work done by worker = F*s * cos(30)

work done by friction = F*s * cos(180) = F*s*-1

I find the friction force by using the frictional coefficient * the normal force

So F = 0.25 * 30 * 9.81 = 73,575N

F*s*cos(30)= -73,575N*s*-1

I can remove s from both sides

F*cos(30) = 73,575N

F=73,575/cos30 = 84,95N

Which is the wrong result

The book states that the answer is 99.2N

What am i doing wrong ?
1. The problem statement, all variables and given/known data



2. Relevant equations



3. The attempt at a solution
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Dr. Jekyll
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#2
Apr21-08, 01:03 PM
P: 35
Of course, the book is right.

Now, you don't need the energies here. It's all about forces. Since the crate moves at constant velocity, the net force on it should be zero (1st Newton's law).

You ignored the y component of the force that the crate is being pushed by. It effects the force that crate exerts on the ground (it also effects the friction!).
astrorob
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#3
Apr21-08, 02:01 PM
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P: 141
As Dr Jekyll states above, leave energies out of it. I can't understand why you're given a distance so ignore that also.

I'd like to say that I can't get that answer if the force acts at an angle below the horizontal yet i get the exact answer if I assume it acts at an angle above, so i'll assume that that part of the information is incorrect (?)

Resolve the components of the force acting on the box into y and x.

Fx = F x cos(30)
Fy = F x sin(30)

The y component of the force acts to increase the normal force, R, experienced by the box;

R = Fy + mg

so you can then work out the frictional force using the formula you've mentioned.

Since the box is at a constant speed, this frictional force should be equal to the x component of force acting on the box. You'll be left with an equation with F. Rearrange this to find your answer.

Hemmelig
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#4
Apr21-08, 02:15 PM
P: 22

Who is right - me or the book - work and kinetic energy


But he's pushing down at the box with an angle of 30 , wont the frictional force be horizontal, as in f x cos (180) then ?
Doc Al
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#5
Apr21-08, 02:24 PM
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P: 40,875
Quote Quote by Hemmelig View Post
Normal force = mg
Since there is an applied force pushing down, the normal force does not equal mg. Figure out the correct normal force by analyzing the vertical components of the forces acting on the crate.
astrorob
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#6
Apr21-08, 02:24 PM
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P: 141
The frictional force is equal to your coefficent x the normal force of the box (that is the force exerted by the ground on the box).

The normal force of the box if he's pushing down on it at an angle of 30 degrees will be greater, that is,

R = Fsin(30) + mg

where Fsin(30) is the component of the force in the y direction.

so frictional force = 0.25 * (Fsin(30) + mg)

Does that make sense?

(Apologies for stepping on your toes Doc Al)


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