How to Calculate Force of Friction and Acceleration for a Block on a Chalkboard?

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Homework Help Overview

The discussion revolves around calculating the force of friction and acceleration for a block being slid up a chalkboard. The problem involves a 0.5 kg block with an applied force of 6.0 N upward and 2.0 N inward toward the board, along with a coefficient of kinetic friction of 0.4.

Discussion Character

  • Mixed

Approaches and Questions Raised

  • Participants explore the relationship between the normal force and the weight of the block, questioning the assumption that the normal force is equal to the weight.
  • There is a discussion about the net force acting on the block in the direction into the board, with some participants suggesting that the acceleration is only upward.
  • Questions arise regarding the implications of a zero net force into the board and how it relates to the forces acting on the block.
  • Some participants suggest considering the forces in two components: upward/downward and into/out of the board.

Discussion Status

The conversation is ongoing, with participants actively questioning assumptions and clarifying concepts related to forces and acceleration. There is a focus on understanding the forces acting on the block and how they relate to the calculations for friction and acceleration.

Contextual Notes

Participants are navigating the complexities of the problem, including the definitions of forces and the implications of Newton's laws. There is a lack of consensus on certain assumptions, particularly regarding the normal force and its role in the calculations.

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Homework Statement


A 0.5kg block is being slid up a chalkboard with an applied force of 6.0N upward and 2.0N inward towards the board. If the coefficient of kinetic friction is 0.4: find the force of friction and the acceleration of the block.


Homework Equations


coefficient of kinetic friction=force of friction/force normal

force of gravity=mass*acceleration

The Attempt at a Solution


for force of friction:

0.4=force of friction/0.5
=0.8

for acceleration:

8N=0.5/acceleration
acceleration=16m/s/s

Is the above method and answer correct? need help.:confused:
 
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You state that the normal force is 0.5? Why?

*Nice username*
 
Isnt force normal same as the weight
\thankyou
 
The normal force is a contact force, it is not the same as the weight. You must always solve for the normal force by using Newton's second law.
The block has a force of 2.0 N into the board. Does the object accelerate into the board?
 
From the question...do u assume that it does??....i am guessing it dosent...only upwards.

Thanks for replying
 
You should picture the motion. It is reasonable to assume that the only acceleration is upwards. So, if the acceleration into the board is zero, what does that imply for the net force into the board?
 
Sorry we took long to reply...we didnt no u posted it...but ne ways...ummmmmm...does the net force into the board equal the applied force?
 
Remember Newtons second law: [tex]F_net = ma[/tex].
So if a is zero in one direction, what can you say about the net force in that direction?
*we?*
 
  • #10
then the net force is 0 isn't it?
 
  • #11
Yes, good.
So what forces act on the block, in the direction of into or out of the board?
 
  • #12
out of the board?
 
  • #13
There is a force acting on the block into the board as mentioned in the problem. You said that the a is zero into the board, hence the net force is zero into the board. If there is one force into the board and the net force is zero...
 
  • #14
so then we do not take the inward force into consideration??/

im confused
 
  • #15
Net force means all the forces. There is stated a force into the board, you said the net force into the board is zero, what does that imply? (are there other forces?)
 
  • #16
isnt there an upward force of 6N though
 
  • #17
Yes, let us just consider the forces into and out of the board for now. Newtons second law is a vector equation which mean you need to only deal with one direction at a time. (Right now we are working on getting the frictional force correct.)
 
  • #18
ok yeah i understand so it means we solve it in two componenets right?
 
  • #19
Yes, an upward/downward direction and an into/out of direction.
You have a force into the board presumably due to the person pushing on the block, are there any other forces in this direction? (or in a direction directly opposite this, which a physicist will just say as the same direction)
 
  • #20
every action there is an equal and opposite reaction?
 
  • #21
So if I push on the block (against the wall) that is a force on the block by me. The block doesn't accelerate into the wall, we can see that physically, so there must be another force on the block (at least one). What could be providing this force?
 
  • #22
The board pushes backwords on us??
 
  • #23
Close, the board does not push back on me, I am not in contact with it, but it does push on something, what is it and how much must it be pushing?
 
  • #24
  • #25
Yes, very nice. So this is the normal force that arises in you friction equation.
 
  • #26
So is the acceleration 0.6??
 
  • #27
Crunching numbers in my head, yes that is it!
 
  • #28
aiit thanks for evrything:D:D
 

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