Why Was My Work Calculation Incorrect?

[physlexic]

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?"

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...

 

[haruspex]

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?"

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.

 

[physlexic]

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 vertical 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.

 

[haruspex]

are you agreeing that I was correct?

Yes.

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. $|\vec x.\vec y| = |\vec x|.|\vec y| |\cos(\theta)|$

 

[physlexic]

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.