Calculate Friction & Acceleration of 2.5kg Block on Horizontal Floor

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A 2.5 kg block is pushed on a horizontal floor with a force of 14 N at an angle of -36°, and the coefficient of kinetic friction is 0.15. To find the frictional force, the equation F_f = μ_k * N is used, but it's noted that the normal force (N) does not equal mg due to the angle of the applied force. A free body diagram (FBD) is recommended to analyze the forces, breaking the applied force into x and y components. The discussion emphasizes the importance of calculating friction before determining acceleration, as the latter depends on the former. Overall, understanding the forces involved and using the correct equations is crucial for solving the problem.
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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 relevent?
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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