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Newtons 2nd Law & Kinetics pt. 2

  • Thread starter jtwitty
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  • #1
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Homework Statement



A 500kg crate is pushed with a 2000N force. The force of the sliding friction acting on the crate is 1900N. How fast will the crate be moving after 10 seconds? How far will it have moved in this time? Determine the net force


Homework Equations



a = [tex]\Sigma[/tex]F/m



The Attempt at a Solution



I tried to find the new force by multiplying 500kg by 10 to get 5000N. Subtracted 5000n frmo 2000n. i don't think negative numbers can be a total force?

2000N-1900N = 100N. ? I don't know what that's used for
 
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Answers and Replies

  • #2
berkeman
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Homework Statement



A 500kg crate is pushed with a 2000N force. The force of the sliding friction acting on the crate is 1900N. How fast will the crate be moving after 10 seconds? How far will it have moved in this time? Determine the net force


Homework Equations



a = [tex]\Sigma[/tex]F/m



The Attempt at a Solution



I tried to find the new force by multiplying 500kg by 10 to get 5000N. Subtracted 5000n frmo 2000n. i don't think negative numbers can be a total force?

2000N-1900N = 100N. ? I don't know what that's used for
Why would you multiple mass by time to get a force?

You are given the two forces on the crate -- what is the "sum" of those forces? Be sure to include direction in the "sum".

And use your equation that you wrote a = [tex]\Sigma[/tex]F/m

to find the acceleration, which is what you need for the rest of the problem.
 
  • #3
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Why would you multiple mass by time to get a force?

You are given the two forces on the crate -- what is the "sum" of those forces? Be sure to include direction in the "sum".

And use your equation that you wrote a = [tex]\Sigma[/tex]F/m

to find the acceleration, which is what you need for the rest of the problem.
No i multiplied by 10 (gravity) (we round up frmo 9.8)
 
  • #4
berkeman
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No i multiplied by 10 (gravity) (we round up frmo 9.8)
You are not given the value of the coefficient of friction mu, so you cannot *calculate* the force of friction opposing the pushing motion. Guess you'll just have to figure out that force in a different way....

(Don't let the forest obscure your view of the trees....) :smile:
 
  • #5
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i don't get what you;re saying hahahah

but ok back to the sum of the forces

2000 + 1900? = 3900

idk
 
  • #6
berkeman
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i don't get what you;re saying hahahah

but ok back to the sum of the forces

2000 + 1900? = 3900

idk
Much closer, but remember that the friction force opposes your pushing force. So if you are pushing with a force vector in the +x direction, what direction is the frictional force vector pointing in? And how do you "add" vectors?
 
  • #7
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Much closer, but remember that the friction force opposes your pushing force. So if you are pushing with a force vector in the +x direction, what direction is the frictional force vector pointing in? And how do you "add" vectors?
idk the word vector :cry:

but i think its pointing in a positive direction?
 
  • #8
berkeman
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idk the word vector :cry:

but i think its pointing in a positive direction?
A vector has magnitude and direction. When two vectors are pointing in opposite directions and you "add" them (like to sum the forces), the resultant vector is the difference in the magnitudes, and points in whichever direction the bigger vector pointed in the first place.

So if you are pushing with 2000N in the +x direction, and the crate's frictional force is pushing in the -x direction with 1900N, what is the resultant vector's magnitude and direction?

You can see the start of this introductory article if you need more info on how to add forces (vectors):

http://en.wikipedia.org/wiki/Vector_addition#Vector_addition_and_subtraction

.
 
  • #9
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it would be poitning in the positive direction b/c its higher by 100N
 
  • #10
berkeman
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it would be poitning in the positive direction b/c its higher by 100N
Bingo! So now you know what the sum of the forces is. Use your equation to calculate the acceleration that results, and then just use the kinematic equations of motion (for constant acceleration) to solve the rest of the problem.
 
  • #11
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Bingo! So now you know what the sum of the forces is. Use your equation to calculate the acceleration that results, and then just use the kinematic equations of motion (for constant acceleration) to solve the rest of the problem.
so r u saying that 100N is the net force?

so a = 100 / 500

a = .2 m/s2
 
  • #12
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help?
 
  • #13
berkeman
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  • #14
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Help why? You are doing fine.
i didn't know if i was right or not :(

was i??
 
  • #15
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a = .2m/s
d = 1/2 (.2) (100)
d = 10

a = vf /t
.2 = vf /10
2 = vf

done? :) how'd i do
 
  • #16
berkeman
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a = .2m/s
d = 1/2 (.2) (100)
d = 10

a = vf /t
.2 = vf /10
2 = vf

done? :) how'd i do
Looks good to me.
 
  • #17
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ty man ty :)
 

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