Why is the work done against friction not equal to the force times distance?

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Homework Help Overview

The discussion revolves around understanding the work done against friction in a physics problem involving a block being moved on a rough horizontal surface. The original poster expresses confusion regarding the relationship between the friction force and the work done against it.

Discussion Character

  • Conceptual clarification, Assumption checking, Problem interpretation

Approaches and Questions Raised

  • Participants explore the definition of the friction force and its role in calculating work done. Questions are raised about why the work done against friction is expressed as f x d rather than a fixed force value of 10N x d.

Discussion Status

Participants are actively engaging with the problem, attempting to clarify the concepts involved. Some have successfully calculated the friction force using Newton's second law, while others are considering different methods to find the further distance traveled by the block after the applied force ceases.

Contextual Notes

The problem involves a block with a specified mass and applied force, and participants are working within the constraints of the information provided, including the acceleration and the time the force is applied.

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Homework Statement


the question and answers are in the attachment, I don't understand question 8b) (ii), why is the work done against friction =f x d and not 10N x d?


Homework Equations





The Attempt at a Solution

 

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For some reason, I cannot view those attachments. Can you describe the problem?
eureka_beyond said:
why is the work done against friction =f x d and not 10N x d?
The obvious question is: What does the friction force equal?
 
okay, the question goes like this:
on a rough horizontal Ground, a 3 kg block at rest is moved be a Force of 10N. The acceleration of the block is 2ms-2.
if the force ceases to act after 4 seconds,
what 's the further distance traveled by the block before it comes to rest?
 
eureka_beyond said:
if the force ceases to act after 4 seconds,
That means the 10 N force is no longer applied.
what 's the further distance traveled by the block before it comes to rest?
What's the only force acting to slow down the block?
 
that's got to be the friction, right?
 
eureka_beyond said:
that's got to be the friction, right?
Yes. Now use the given information to figure out the friction force. (Hint: Newton's 2nd law.)
 
by Newton's 2nd law
F=ma
Let the friction be f
10-f=3(2)
f=4N
 
eureka_beyond said:
by Newton's 2nd law
F=ma
Let the friction be f
10-f=3(2)
f=4N
Good!
 
yay! thanks a million. but there's another thing, how do I get the further distance traveled by the block before it comes to rest?
 
  • #10
eureka_beyond said:
yay! thanks a million. but there's another thing, how do I get the further distance traveled by the block before it comes to rest?
Several ways:
(1) You can find the acceleration and use kinematics. (First figure out how fast the block is moving at the end of 4 seconds.)
(2) You can set the work done by friction equal to the change in kinetic energy.
 
  • #11
ah, I see. The book must have been using the second way you describe, I get it now, thanks a lot.
 

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