Solving a Homework Question: 25N Force Opposite to Displacement?

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Homework Help Overview

The discussion revolves around a physics homework problem involving a 25N force acting opposite to the direction of displacement on a block moving along a frictionless surface. The original poster expresses confusion about the direction of the force and its implications for calculating work done.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the calculation of work done using the formula W=(F cos θ) s, with some questioning the angle used in the calculation due to the force's direction. There is also a focus on understanding the implications of a negative acceleration and the nature of the work done.

Discussion Status

Some participants provide guidance on the correct angle to use for the calculation, suggesting that the angle should be adjusted to account for the force acting opposite to the displacement. The conversation reflects a mix of interpretations regarding the problem's setup and the calculations involved.

Contextual Notes

The original poster notes uncertainty about the problem's difficulty level, indicating a belief that it may be simpler than typical assignments from their professor. There is also mention of a lack of response from the teacher, which adds to the original poster's confusion.

dragon84
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I have a homework question that has a box and it has the 25N force acting on it in the opposite direction of the displacement. Is this possible or is the force poining in the wrong direction. I email my teacher but he hasnt replied yet.
You can view a diagram i made
HERE
 
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I don't see the problem... The acceleration will simply be negative, that is, the body will slow down. What's bothering you?
 
I don't know what's bothering me, it just does. I am also not looking for acceleration. I am looking for W.

the problem says-

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

do I just use W=(F cos [tex]\theta[/tex]) s
.....W=(25 cos 30) 7.5=162.37J

If that's it(which I doubt b/c I don't know what I am talking about when it come to physcs), its way too easy of a prob that my professor usually gives out for HW
 
You do just that, only the angle is now 210 instead of 30, that's all. (x's direction is to the right, so the angle with that direction isn't 30 but 180+30=210).
So you'll get exactly the negative value you got, which would mean to body does work on the guy that applies the force... (For instance, it could accelerate something that would be attached to it).
 
Ok, thanks for helping me out with that. I guess it was hard for me to understand b/c I was looking into it too much b/c my professor never gives out easy homework problems.
 

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