Calculate Friction & Acceleration of 2.5kg Block on Horizontal Floor

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Homework Help Overview

The discussion revolves around calculating the frictional force and acceleration of a 2.5 kg block being pushed on a horizontal floor. The problem involves a force applied at an angle and includes the coefficient of kinetic friction.

Discussion Character

  • Exploratory, Conceptual clarification, Mathematical reasoning, Problem interpretation

Approaches and Questions Raised

  • Participants discuss relevant equations and the need for a free body diagram to analyze the forces acting on the block. There are attempts to break down the applied force into components and questions about the normal force.

Discussion Status

Some participants are exploring different interpretations of the forces involved, particularly the normal force and its relationship to the applied force. Guidance has been offered regarding the need to express equations symbolically and the importance of determining the frictional force before calculating acceleration.

Contextual Notes

There is uncertainty regarding the normal force due to the angle of the applied force, and some participants express a lack of understanding about the necessary equations and concepts involved.

randybrent
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Friction No Clue!

A 2.5 kg block is pushed along a horizontal floor by a force of magnitude 14 N at an angle = -36° with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.15. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

Any help would be welcomed.
 
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C'mon, you have to have some clue!
What equations might be relevant?
What do you know?
What are you looking for?
 
14n cos 36 - .15 =11.17 n ?

11.17/3.5= 3.19 m/s^2 ?
 
First of all, no.
Second of all, I meant equations as in symbols.
like this:

v=x/t
x=123
t=.456

But yours is trickier. But not impossible:wink: .
 
randybrent said:
A 2.5 kg block is pushed along a horizontal floor by a force of magnitude 14 N at an angle = -36° with the horizontal. The coefficient of kinetic friction between the block and the floor is 0.15. Calculate the magnitudes of (a) the frictional force on the block from the floor and (b) the block's acceleration.

Any help would be welcomed.

It helps to draw a FBD(free body diagram) of the block and write all the forces acting on it. I see 4 different forces, can you? (One of the force is at an angle, so that can be broken up into x and y components)
 
w
i
l
l
uk -------------------- 14 cos 36
l
l
l
mg 14 sin 36

does this look right?
 
Ummm... I don't get it.

About the equations...
you're looking for the force of friction.
The force of kinetic friction is given by the equaiton F_f=\mu_kN
What do you know?
What are you looking for?
What do you still need to know?
Is there another equation you could use to find that?
 
i don't know enough i guess
 
Ff= .15*2.5kg*9.8= 3.675n ?
 
  • #10
OK, you're getting there.
Unfortunately, in this case N doesn't equal mg, even though it often does.
Use \Sigma F_y=ma=0, and notice that the force of the push has a y component.
 
  • #11
a= [.15*14sin36*9.8*2.5] /2.5 ?
 
  • #12
Look, Randy, can you do me a favor?
Don't just show me the numbers; state the equation symbolically first.

Also, I highly suggest that you solve for friction before looking for acceleration--you can't find acceleration without knowing the force of friction.
 
  • #13
i give up thank you for your time
 

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