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Newton's Law and Forces

  • Thread starter Kmcquiggan
  • Start date
Problem Statement
A child pushes a block of wood with a mass of 0.72kg across a smooth table. The block starts from a position of rest and after 2.0 s it has a velocity of 1.6m/s [forward]. The coefficient of kinetic friction is 0.64
Relevant Equations
FN= Fg=mg , Fnet=Fg+FN, Ff=Fnuk
I am very new to physics so I am still learning a lot. Here is my attempt:
Find the net force acting on the block : Fnet= Fg+FN so I have to find FN before I can complete the answer. FN = Fg (mg) FN = (0.72kg)(9.8m/s^2) = 7.056 or 7.06 N
Fnet = 9.8+7.06 = 16.86 N or 16.9 N

To find the force of friction I used the formula Ff= (FN)(uk) = 16.9*.64 =10.816N or 10.8 N

Now I used the formula Fapplied = Ff+FN = 10.8N + 7.06N = 17.86 or 17.9 N of applied force.

Am I even close to understanding this or correct?
 

neilparker62

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$$ F_{net}=ma $$ so you can determine ## F_{net}##. Then: $$ F_{net}=F_{applied}-F_{friction} $$ with $$ F_{friction}=μ_kmg $$
 
ok thank you, so i was way off lol
 

PeroK

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ok thank you, so i was way off lol
Have you ever heard the term "free-body diagram"? If not, do an Internet search.

Two questions:

1) How many forces are acting on the block?

2) What is the relationship between net force and acceleration?
 
$$ F_{net}=ma $$ so you can determine ## F_{net}##. Then: $$ F_{net}=F_{applied}-F_{friction} $$ with $$ F_{friction}=μ_kmg $$
For the first formula are you meaning FN? Just wondering as you have 2 formulas for Fnet? Or am I missing something?
 
Ok, so with the new understanding
Fnet = ma or Fnet= (0.72)(9.8m/s^2) = 7.056 or 7.06N
Ff= Ukmg = (0.64)(.72)(9.8) = 4.51584 or 4.52 N
The formula I was given to find Fnet (horizontal) = Fapplied +Ff so to find applied force can I take Fnet- Ff and that will give me applied? 7.06-4.52 = 2.54N for applied force?
 

PeroK

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Ok, so with the new understanding
Fnet = ma or Fnet= (0.72)(9.8m/s^2) = 7.056 or 7.06N
Ff= Ukmg = (0.64)(.72)(9.8) = 4.51584 or 4.52 N
The formula I was given to find Fnet (horizontal) = Fapplied +Ff so to find applied force can I take Fnet- Ff and that will give me applied? 7.06-4.52 = 2.54N for applied force?
That's not correct, I'm sorry to say.

To start at the beginning. You say:

Fnet = ma or Fnet= (0.72)(9.8m/s^2) = 7.056 or 7.06N

Why do you think the acceleration of the block is ##9.8 m/s^2##?
 

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