Why Was My Work Calculation Incorrect?

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In summary, the conversation discusses a problem on a physics exam where the wrong formula was used to find the work done by a constant force on a block undergoing displacement. The person initially used cosine but the correct answer was found using sine. They question why their professor did not clarify this and plan to confront them about it. The expert suggests that it depends on how the angle is defined and gives examples to support using sine in certain situations. Overall, the person agrees with the expert's explanation.
  • #1
physlexic
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Homework Statement

On my exam, I got this problem wrong, which I believed was incredibly easy, and I don't know why it's wrong...

"A constant force of 25 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force?"
Juhnd9L.jpg


Homework Equations


W = F * S

The Attempt at a Solution


There is a displacement to the right, while the force vector is to the left, therefore work is in the negative.

displacement (s) = 7.5 m
F = 25 N (cos 30 degrees)

I chose cosine because we are only concerned for the x direction...which is adjacent to the angle of 30 degrees...

therefore
W = F * s = [25 N (cos 30)] * [-7.5 m]
W = -162 J

However the answe was -94 J...and this is because sine was used instead. Why? My professor didn't care to explain why...
 
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  • #2
physlexic said:

Homework Statement

On my exam, I got this problem wrong, which I believed was incredibly easy, and I don't know why it's wrong...

"A constant force of 25 N is applied as shown to a block which undergoes a displacement of 7.5 m to the right along a frictionless surface while the force acts. What is the work done by the force?"
Juhnd9L.jpg


Homework Equations


W = F * S

The Attempt at a Solution


There is a displacement to the right, while the force vector is to the left, therefore work is in the negative.

displacement (s) = 7.5 m
F = 25 N (cos 30 degrees)

I chose cosine because we are only concerned for the x direction...which is adjacent to the angle of 30 degrees...

therefore
W = F * s = [25 N (cos 30)] * [-7.5 m]
W = -162 J

However the answe was -94 J...and this is because sine was used instead. Why? My professor didn't care to explain why...
I can only suggest you ask your professor what the answer would have been with an angle of 90 degrees.
 
  • #3
are you agreeing that I was correct? because if so that's a whole letter grade I get back to my test if I can stump my professor and prove that I was right...and I help out my classmates with the same answer as me.

and that's true...because if it was 90 degrees no work would be done but since he used sine which equals 1 it's implying work was done...and a verticle force never causes a horizontal displacement.

I thought you also never use sine for work. I never came across a problem where you had to besides this one.
 
  • #4
physlexic said:
are you agreeing that I was correct?
Yes.
physlexic said:
I thought you also never use sine for work.
It depends how you are defining the angle. What if F were dragging the block up against a vertical wall? Or if the given angle were measured against the vertical?
If you are defining the angle to be that between the force vector and the displacement vector then I agree. ##|\vec x.\vec y| = |\vec x|.|\vec y| |\cos(\theta)|##
 
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  • #5
haruspex said:
Yes.

It depends how you are defining the angle. What if F were dragging the block up against a vertical wall? Or if the given angle were measured against the vertical?
If you are defining the angle to be that between the force vector and the displacement vector then I agree. ##|\vec x.\vec y| = |\vec x|.|\vec y| \cos(\theta)##

I see from your examples, it would make sense to use sine then in those situations.

thank you for the explanation. I suppose I will confront him about this, as nice as possible.
 

FAQ: Why Was My Work Calculation Incorrect?

What is work?

Work is a physical quantity that measures the amount of energy used to move an object over a distance. It is calculated by multiplying the force applied to an object by the distance the object moves.

How is work related to force?

Work is directly proportional to force, meaning that as the force increases, the amount of work done also increases. This can be seen in the work formula: W = F * d, where W represents work, F represents force, and d represents distance.

What is the unit of measurement for work?

The unit of measurement for work is the joule (J). One joule is equal to the amount of work done when a force of one newton is applied to an object and moves it one meter in the direction of the force.

How do you calculate the work done by a force?

The work done by a force can be calculated by multiplying the magnitude of the force by the distance the object moves in the direction of the force. This can be represented by the formula W = F * d.

Can work be negative?

Yes, work can be negative. This occurs when the force and the object's displacement are in opposite directions. This means that the force is doing work in the opposite direction of the object's motion, resulting in a negative value for work.

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