Work and Energy Question ( by 11 PM tonight if possible please)

[kasiu]

1. A 2.00 x10^2-N force is pulling an 75.0-kg refrigerator across a horizontal surface. The force acts at an angle of 23.0° above the surface. The coefficient of kinetic friction is 0.200, and the refrigerator moves a distance of 9.00 m. Find (a) the work done by the pulling force, and(b) the work done by the kinetic frictional force.

I need help with part b of this problem
mass: 75 kg
distance: 9 m
Force: 200
theta: 23 degrees


2. Work= Force*distance* Cos(theta)
Friction=mu*Normal Force



3. (mass*9.8)-Tension*Sin(theta)=Normal Force
((mass*9.8)-tension*Sin(theta))*mu=Friction Force
Friction Force*distance=work done by frictional force

(75*9.8) - (200*sin(23)) = Fn
735-78.146= Fn
656.85377 = Fn
so,
656.85377*.2=Friction force
131.371= Friction force
so,
131.371*9=work
1182.337 J = work

I have a few questions:
First, what am I doing wrong?
Is cos(theata) = cos(0) = 1
and
would the final answer for work be negative because this is the work of friction?

Any help would be appreciated greatly!
Thank you ahead of time :)

 

[cepheid]

I have a few questions:
First, what am I doing wrong?

Kind of a strange question to ask for two reasons.
1. We don't have the correct answer or the solution in front of us, so we can't tell you what's wrong with your solution unless we go through it line by line and happen to find a computational mistake (or a mistake in the physics).

2. You haven't said anything to indicate how it is that you know that something is wrong.

Is cos(theata) = cos(0) = 1

Theta is 23 degrees. You said so yourself. Multiplying the pulling force by cos(theta) gives you the horizontal component of that force (the component that actually does work).

would the final answer for work be negative because this is the work of friction?

Yes, it would be negative because the direction of the frictional force is *opposite* to the direction of the displacement of the fridge.

 

[Doc Al]

I have a few questions:
First, what am I doing wrong?

Nothing.

Is cos(theata) = cos(0) = 1
and
would the final answer for work be negative because this is the work of friction?

Yes, the answer should be negative because friction and displacement are in opposite directions. (If you like, you can treat the angle between them as 180 degrees. Cos(180) = -1.)

 

[kasiu]

So, quick question:
in my final formula to find work of the kinetic
would i set it up as
Work = Force*cos(theata)*distance
Work= 131.371*cos(23)*9
or would i use
Work=131.371*cos(0)* 9
because cosine would be 0 degrees on the horizontal?

 

[kasiu]

Oh!
Thank you very much!
The both of you have been very helpful!
Once again,
thank you!
:)

 

[Doc Al]

I'm sure you have it now, but just in case:

So, quick question:
in my final formula to find work of the kinetic
would i set it up as
Work = Force*cos(theata)*distance
Work= 131.371*cos(23)*9
or would i use
Work=131.371*cos(0)* 9
because cosine would be 0 degrees on the horizontal?

You would use theta = 0 (or 180) degrees. 23 degrees is the angle of the applied force, not the friction, so it's not directly relevant for part b.

 

[cepheid]

Yeah, my bad. I didn't realize you were referring to theta for the frictional force, and not for the applied force. Anyway, it looks like Doc Al sorted it out.